5.1 The sign of work done by a force on a body is important to understand. State carefully. if the following quantities are positive or negative:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Solution :
(a) Positive: The man applies force upwards, and the bucket moves upwards. Force and displacement are in the same direction, so the work done is positive.
(b) Negative: Gravitational force acts downwards, while the bucket moves upwards. Force and displacement are in opposite directions, so the work done is negative.
(c) Negative: Frictional force acts opposite to the direction of motion of the body sliding down the inclined plane. Force and displacement are in opposite directions, so the work done is negative.
(d) Positive: The applied force acts in the direction of motion of the body on the horizontal plane. Force and displacement are in the same direction, so the work done is positive.
(e) Negative: The resistive force of air acts opposite to the direction of motion of the pendulum. Force and displacement are in opposite directions, so the work done is negative.
5.2 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.Compute the
(a) work done by the applied force in 10 s,
(b) work done by friction in 10 s,
(c) work done by the net force on the body in 10 s,
(d) change in kinetic energy of the body in 10 s, and interpret your results.
Solution:
First, let's calculate the acceleration of the body:
* Given:
* Mass (m) = 2 kg
* Applied force (F) = 7 N
* Coefficient of kinetic friction (μ) = 0.1
* Initial velocity (u) = 0 m/s (initially at rest)
* Time (t) = 10 s
* Calculate frictional force:
* Frictional force (f) = μ * m * g
* f = 0.1 * 2 * 9.8 = 1.96 N
* Calculate net force:
* Net force (Fnet) = F - f = 7 - 1.96 = 5.04 N
* Calculate acceleration:
* Acceleration (a) = Fnet / m = 5.04 / 2 = 2.52 m/s²
Now, let's calculate the distance traveled in 10 seconds:
* Distance (s) = ut + 0.5at²
* s = 0 + 0.5 * 2.52 * 10² = 126 m
Now, we can calculate the work done by each force:
(a) Work done by the applied force:
* Work (W) = Force × Distance
* W_applied = F × s = 7 N × 126 m = 882 J
(b) Work done by friction:
* Work (W) = Force × Distance
* W_friction = f × s = 1.96 N × 126 m = -247.08 J (negative sign indicates work done against the force)
(c) Work done by the net force:
* Work (W) = Force × Distance
* W_net = Fnet × s = 5.04 N × 126 m = 634.56 J
(d) Change in kinetic energy:
* Initial kinetic energy (KE_initial) = 0.5 * m * u² = 0 J (initially at rest)
* Final velocity (v) = u + at = 0 + 2.52 * 10 = 25.2 m/s
* Final kinetic energy (KE_final) = 0.5 * m * v² = 0.5 * 2 * (25.2)² = 634.56 J
* Change in kinetic energy = KE_final - KE_initial = 634.56 J - 0 J = 634.56 J
Interpretation:
* The work done by the net force is equal to the change in kinetic energy of the body. This confirms the work-energy theorem.
* The work done by the applied force is positive, indicating that the force adds energy to the system.
* The work done by friction is negative, indicating that friction removes energy from the system, converting it into heat.
*
5.3 Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.
Solution:
(a) The kinetic energy of a body cannot be negative because it is a positive quantity.
From the above graph it can be observed that V 0 >0 for the region x>a.
The kinetic energy of the system is given as,
K.E=E−P.E
From the graph, it can be observed that the potential energy is greater than total energy for the given region so the kinetic energy will be negative and it is not possible. The particle will not exist in that region.
Thus, the particle will not exist in this region and the minimum total energy of the particle will be zero.
(b) From the above graph, it can be observed that the potential energy is always ( −∞>x>+∞ ) greater than the total energy. So, the kinetic energy will be negative and it is not possible for the particle to be exists in this region.
c)From the graph, it is clear that the potential energy is − V 1 for the region x>a and x<b and potential energy is V 0 for the region x<a and x>b.
The condition for the kinetic energy to be positive is satisfied only for the region x>a and x<b
K.E=E−P.E =E−( − V 1 ) =E+ V 1
Thus, the minimum total energy is − V 1 and the region is region x>a and x<b.
(d) The condition for kinetic energy to be positive, the region −b 2 <x< b 2 and −a 2 <x< a 2 will be satisfied region.
The minimum potential energy is − V 1 .
The kinetic energy of the system is given as,
K.E=E−P.E =E−( − V 1 ) =E+ V 1
Thus, the minimum total energy is − V 1 and the region is −b 2 <x< b 2 and −a 2 <x< a 2 .
5.4 The potential energy function for a particle executing linear simple harmonic motion is given by $V(x) = kx^{2}/2$, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.
Solution:
We are given the potential energy function:
V(x) = (1/2)kx^2
and the total energy E = 1 J, and k = 0.5 N/m.
A particle will "turn back" when its kinetic energy becomes zero, meaning all its energy is potential energy. So, we can set the potential energy equal to the total energy and solve for x:
1 J = (1/2)(0.5 N/m) * x^2
Multiplying both sides by 2 and dividing by 0.5 N/m:
4 m^2 = x^2
Taking the square root of both sides:
x = ±2 m
Therefore, the particle will turn back when it reaches
x = ±2 m.
5.5 Answer the following :
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why ?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ?
(d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig.5.13 (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ?
Solution:
(a) The rocket's energy. The heat energy required for burning the casing is obtained from the chemical energy stored in the rocket's fuel. As the fuel burns, it releases energy, a portion of which is used to accelerate the rocket and another portion is used to heat and burn the casing.
(b) Gravitational force is a conservative force. This means that the work done by a conservative force on an object moving along a closed path is zero. In the case of a comet orbiting the sun, its elliptical orbit is a closed path. Therefore, the total work done by the gravitational force over one complete orbit is zero.
(c) As the satellite comes closer to the Earth, its potential energy decreases. This decrease in potential energy is converted into kinetic energy, causing the satellite's speed to increase. While atmospheric resistance does cause a gradual loss of energy, the effect of decreasing potential energy is more significant in this case.
(d) The work done is greater in Fig. 5.13(ii). In the first case, the man is only doing work against gravity to lift his own body. In the second case, he is doing work against gravity to lift both his body and the 15 kg mass. Since the total weight being lifted is greater in the second case, the work done is also greater.
5.6 Underline the correct alternative :
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
Solution:
(a) When a conservative force does positive work on a body, the potential energy of the body decreases.
(b) Work done by a body against friction always results in a loss of its kinetic energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total linear momentum of the system of two bodies.
5.7 State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
Solution:
(a) False. In an elastic collision, the total kinetic energy and total momentum of the system are conserved, not necessarily for each individual body.
(b) False. Total energy is conserved only when there are no non-conservative forces acting on the system. Non-conservative forces like friction can dissipate energy as heat, sound, etc., leading to a decrease in the total mechanical energy of the system.
(c) False. This is only true for conservative forces. Non-conservative forces like friction can do non-zero work over a closed loop. For example, if you push a block along a rough surface in a closed loop, the work done by friction will not be zero.
(d) True. In an inelastic collision, some kinetic energy is lost to other forms of energy, such as heat or sound. Therefore, the final kinetic energy is always less than the initial kinetic energy of the system.
5.8 Answer carefully, with reasons :
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact) ?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls ?
(c) What are the answers to (a) and (b) for an inelastic collision ?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic ? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
Solution :
(a) No. During the short time of collision, the kinetic energy is converted into potential energy as the balls deform. This potential energy is then converted back into kinetic energy as the balls separate. However, during the deformation phase, the total mechanical energy is not purely kinetic.
(b) Yes. Linear momentum is conserved during the entire collision process, including the short time of contact. This is because there are no external forces acting on the system of two balls during the collision.
(c) For an inelastic collision:
* Kinetic energy: Not conserved. Some kinetic energy is lost to other forms of energy, such as heat or sound.
* Linear momentum: Conserved. Linear momentum is conserved in all types of collisions, elastic or inelastic.
(d) Elastic. If the potential energy depends only on the separation distance, it means that the force between the balls is conservative. A conservative force leads to elastic collisions, where kinetic energy is conserved.
5.9 A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(i) $t^{1/2}$ (ii) t (iii) $t^{3/2}$ (iv) $t^{2}$
Solution:
To determine the power delivered to a body undergoing one-dimensional motion with constant acceleration, we can use the following steps:
Velocity Calculation: Since the body starts from rest, its initial velocity ( u = 0 ). With constant acceleration ( a ), the velocity ( v ) at time ( t ) is given by:
[ v = u + at = at ]
Force Calculation: The force ( F ) acting on the body is: [ F = ma ] where ( m ) is the mass of the body.
Power Calculation: Power ( P ) is the product of force and velocity:
$[ P = F \cdot v = ma \cdot at = ma^2 t ]$
Since ( m ) and ( a ) are constants, the power ( P ) is directly proportional to ( t ): [ $P \propto t$ ]
Therefore, the correct answer is (ii) ( t ).
5.10 A body is moving unidirectionally under the influence of a source of constant power.Its displacement in time t is proportional to
(i) $t^{1/2}$ (ii) t (iii) $t^{3/2}$ (iv) $t^{2}$
Solution:
When a body moves under the influence of a constant power source, its displacement (d) in time (t) is proportional to $(t^{3/2})$. This relationship can be derived from the equations of motion and the definition of power. Here’s a brief outline of the reasoning:
Power (P) is defined as the rate at which work is done, ($P = \frac{W}{t}$), where (W) is the work done.
Work done (W) is the product of force (F) and displacement (d), ($W = F \cdot d$).
For a body of mass (m), moving with acceleration (a), the force (F) is given by ($F = m \cdot a$).
The power can also be expressed as ($P = F \cdot v$), where (v) is the velocity.
Substituting ($F = m \cdot a$) and ($v = \frac{d}{t}$) into the power equation, we get ($P = m \cdot a \cdot \frac{d}{t}$).
Since power is constant, ($P = \text{constant}$), we can solve for (d) in terms of (t), leading to ($d \propto t^{3/2}$).
Thus, the displacement (d) is proportional to ($t^{3/2}$).
5.11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force $\vec{F}$ given by
$\vec{F} = -\hat{i} + 2\hat{j} + 3\hat{k} \text{ N}$
where $\hat{i}$, $\hat{j}$, and $\hat{k}$ are unit vectors along the x-, y-, and z-axis of the system, respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?
Solution:
Understanding the Problem:
We're given a force vector:
$\vec{F} = -\hat{i} + 2\hat{j} + 3\hat{k} N$
And a displacement vector along the z-axis:
$\vec{d} = 4\hat{k} m$
Calculating Work Done:
Work done by a constant force is the dot product of the force and displacement vectors:
$W = \vec{F} \cdot \vec{d}$
Substituting the given values:
$W = (-\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (4\hat{k})$
Using the properties of the dot product:
$W = (-1)(0) + (2)(0) + (3)(4)$
Calculating the result:
$W = 12 J$
Therefore, the work done by the force in moving the body 4 meters along the
z-axis is 12 Joules.
5.12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10-31 kg, proton mass= 1.67×10–27 kg, 1 eV = 1.60 ×10–19 J).
Solution:
Let's analyze the problem:
Given:
* Mass of electron, mₑ = 9.11 × 10⁻³¹ kg
* Mass of proton, mₚ = 1.67 × 10⁻²⁷ kg
* Kinetic energy of electron, KEₑ = 10 keV = 10 × 10³ × 1.6 × 10⁻¹⁹ J
* Kinetic energy of proton, KEₚ = 100 keV = 100 × 10³ × 1.6 × 10⁻¹⁹ J
To find:
* Which particle is faster?
* Ratio of their speeds.
Solution:
1. Kinetic Energy and Speed:
The kinetic energy of a particle is given by the formula:
KE = 0.5 * m * v²
where:
* KE is the kinetic energy
* m is the mass of the particle
* v is the speed of the particle
2. Speed of Electron:
Rearranging the formula for speed:
vₑ = √(2 * KEₑ / mₑ)
Substituting the values:
vₑ = √(2 * 10 × 10³ × 1.6 × 10⁻¹⁹ J / 9.11 × 10⁻³¹ kg)
Calculating this gives:
vₑ ≈ 5.93 × 10⁷ m/s
3. Speed of Proton:
Similarly, for the proton:
vₚ = √(2 * KEₚ / mₚ)
Substituting the values:
vₚ = √(2 * 100 × 10³ × 1.6 × 10⁻¹⁹ J / 1.67 × 10⁻²⁷ kg)
Calculating this gives:
vₚ ≈ 4.38 × 10⁶ m/s
4. Comparing Speeds and Ratio:
* Electron is faster than the proton.
* Ratio of speeds:
vₑ / vₚ = (5.93 × 10⁷ m/s) / (4.38 × 10⁶ m/s) ≈ 13.5
Conclusion:
The electron, with a smaller mass and higher kinetic energy, is moving significantly faster than the proton. The electron's speed is
approximately 13.5 times the proton's speed.
5.13 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s–1 ?
Solution:
Analyzing the Rain Drop's Journey
Work Done by Gravity
First Half:
* The rain drop falls from 500 m to 250 m.
* Work done by gravity = mgh
* m = mass of the raindrop
* g = acceleration due to gravity (9.81 m/s²)
* h = vertical distance (250 m)
Second Half:
* The rain drop falls from 250 m to 0 m.
* Work done by gravity = mgh
* m = mass of the raindrop
* g = acceleration due to gravity (9.81 m/s²)
* h = vertical distance (250 m)
Note: In both halves, the work done by gravity is the same because the vertical distance traveled is the same.
Work Done by Resistive Force
The total work done on the raindrop is equal to its change in kinetic energy.
* Initial potential energy = mgh (at 500 m)
* Final kinetic energy = 0.5mv² (at ground level)
* Work done by resistive force = Initial potential energy - Final kinetic energy
Calculating the Mass of the Raindrop:
* Volume of the raindrop = (4/3)πr³
* Mass of the raindrop = Volume × Density of water
Calculating the Values:
* Mass of the raindrop:
* Volume = (4/3)π(2 × 10⁻³ m)³ ≈ 3.35 × 10⁻⁸ m³
* Mass = (3.35 × 10⁻⁸ m³) × (1000 kg/m³) ≈ 3.35 × 10⁻⁵ kg
* Work done by gravity in each half:
* Work = (3.35 × 10⁻⁵ kg)(9.81 m/s²)(250 m) ≈ 0.082 J
* Total work done by gravity:
* Total work = 2 × 0.082 J ≈ 0.164 J
* Initial potential energy:
* Initial PE = (3.35 × 10⁻⁵ kg)(9.81 m/s²)(500 m) ≈ 0.164 J
* Final kinetic energy:
* Final KE = 0.5(3.35 × 10⁻⁵ kg)(10 m/s)² ≈ 0.001675 J
* Work done by resistive force:
* Work = 0.164 J - 0.001675 J ≈ 0.162 J
Therefore:
* Work done by gravity in the first half: 0.082 J
* Work done by gravity in the second half: 0.082 J
* Total work done by gravity: 0.164 J
* Work done by resistive force: 0.162 J
5.14 A molecule in a gas container hits a horizontal wall with speed 200 m s–1 and angle 30°with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ?
Solution:
Momentum Conservation:
* Yes, momentum is conserved in the collision.
The key to understanding this is to consider the system as the molecule and the wall. While the molecule's momentum changes direction during the collision, the wall also experiences a change in momentum. These changes are equal and opposite, ensuring that the total momentum of the system remains constant.
Elastic or Inelastic Collision:
* The collision is elastic.
In an elastic collision, both kinetic energy and momentum are conserved. In this case, the molecule rebounds with the same speed, indicating that its kinetic energy remains unchanged. Therefore, the collision is elastic.
5.15 A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%,how much electric power is consumed by the pump ?
Solution:
Here's how to calculate the electric power consumed by the pump:
1. Calculate the work done by the pump:
* The pump lifts a volume of 30 m³ of water to a height of 40 m.
* The mass of this water is:
mass = density × volume = 1000 kg/m³ × 30 m³ = 30000 kg
* The work done to lift this mass against gravity is:
work = mgh = 30000 kg × 9.8 m/s² × 40 m = 11760000 J
2. Calculate the power output of the pump:
* The pump does this work in 15 minutes, which is 900 seconds.
* So, the power output of the pump is:
power_output = work / time = 11760000 J / 900 s ≈ 13067 W
3. Calculate the electric power input to the pump:
* The efficiency of the pump is 30%, which means that only 30% of the electric power input is converted into useful work.
* So, the electric power input is:
electric_power_input = power_output / efficiency = 13067 W / 0.3 ≈ 43557 W
Therefore, the electric power consumed by the pump is approximately 43.6 kW.
5.16 Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision ?
Solution:
Solution :
Understanding the Problem:
* We have three identical balls: A, B, and C.
* Initially, only ball A is moving with velocity V.
* Balls B and C are at rest and touching each other.
* The collision is elastic, meaning both kinetic energy and momentum are conserved.
Analyzing the Options:
Let's analyze each option to see if it conserves both kinetic energy and momentum:
Option (i): Ball A stops, and balls B and C move with velocity V/2 each.
* Momentum Conservation:
* Initial momentum = mV
* Final momentum = m(V/2) + m(V/2) = mV
* Momentum is conserved.
* Kinetic Energy Conservation:
* Initial KE = 1/2 mV²
* Final KE = 1/2 m(V/2)² + 1/2 m(V/2)² = 1/4 mV²
* Kinetic energy is NOT conserved.
Option (ii): Ball A stops, and only ball C moves with velocity V.
* Momentum Conservation:
* Initial momentum = mV
* Final momentum = mV
* Momentum is conserved.
* Kinetic Energy Conservation:
* Initial KE = 1/2 mV²
* Final KE = 1/2 mV²
* Kinetic energy is conserved.
Option (iii): All three balls move with velocity V/3.
* Momentum Conservation:
* Initial momentum = mV
* Final momentum = 3m(V/3) = mV
* Momentum is conserved.
* Kinetic Energy Conservation:
* Initial KE = 1/2 mV²
* Final KE = 3 * 1/2 m(V/3)² = 1/6 mV²
* Kinetic energy is NOT conserved.
Conclusion:
Option (ii) is the only possible outcome after the collision because it conserves both momentum and kinetic energy. In this scenario, ball A transfers all its momentum and kinetic energy to ball C, leaving itself at rest.
5.17 The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic.
Solution:
Understanding the Problem:
In this problem, we have an elastic collision between two identical masses. The initial velocity of bob A can be determined using conservation of energy. After the collision, momentum and kinetic energy are conserved.
Solution:
* Initial Velocity of Bob A:
* When released from 30 degrees, bob A loses potential energy and gains kinetic energy.
* At the lowest point, all potential energy is converted to kinetic energy.
* Using conservation of energy:
mgh = 0.5mv^2
* Where:
* m is the mass of the bob
* g is the acceleration due to gravity (9.81 m/s^2)
* h is the vertical height loss (L - Lcos30°, where L is the pendulum length)
* v is the velocity at the lowest point
* Solving for v:
v = sqrt(2gL(1 - cos30°))
* Elastic Collision:
* In an elastic collision between equal masses, the incoming mass comes to rest, and the target mass moves with the initial velocity of the incoming mass.
* So, after the collision, bob B moves with velocity v, and bob A comes to rest.
* Final Height of Bob A:
* Bob A swings back up, converting its kinetic energy back to potential energy.
* Using conservation of energy again:
0.5mv^2 = mgh'
* Where h' is the final height of bob A.
* Solving for h':
h' = v^2 / (2g)
* Substituting the value of v from step 1:
h' = (2gL(1 - cos30°)) / (2g)
* Simplifying:
h' = L(1 - cos30°)
Therefore, the bob A rises to the same height from which it was released, which is 30 degrees above the vertical.
5.18 The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m,what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance ?
Solution:
Here's how to solve the problem:
* Calculate the initial potential energy:
* When the bob is released from a horizontal position, all its energy is in the form of potential energy.
* Potential energy (PE) = mgh, where m is the mass of the bob, g is the acceleration due to gravity (9.8 m/s^2), and h is the initial height.
* Since the bob is released from a horizontal position, the initial height h is equal to the length of the pendulum (l): h = 1.5 m.
* So, PE = m * g * 1.5.
* Calculate the final kinetic energy:
* 5% of the initial energy is dissipated, so 95% of the initial energy is converted into kinetic energy at the lowest point.
* Kinetic energy (KE) = 0.95 * PE = 0.95 * m * g * 1.5.
* Calculate the final speed:
* Kinetic energy is also given by KE = 0.5 * m * v^2, where v is the final speed.
* Equating the two expressions for kinetic energy:
* 0.95 * m * g * 1.5 = 0.5 * m * v^2.
* The mass m cancels out:
* 0.95 * g * 1.5 = 0.5 * v^2.
* Solving for v:
* v = sqrt(2 * 0.95 * g * 1.5) ≈ 5.3 m/s.
Therefore, the speed with which the bob arrives at the lowermost point is approximately 5.3 m/s.
5.19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s–1. What is the speed of the trolley after the entire sand bag is empty ?
Solution:
Here's how to solve the problem:
Step 1: Convert the initial velocity to m/s:
Initial velocity (v1) = 27 km/h = 27 * (1000 m/km) / (3600 s/h) = 7.5 m/s
Step 2: Calculate the initial momentum:
Initial mass (m1) = mass of trolley + mass of sandbag = 300 kg + 25 kg = 325 kg
Initial momentum (p1) = m1 * v1 = 325 kg * 7.5 m/s = 2437.5 kg m/s
Step 3: Calculate the final mass:
When the sandbag is empty, the final mass of the system is just the mass of the trolley:
Final mass (m2) = 300 kg
Step 4: Apply the conservation of momentum:
Since there are no external forces acting on the system in the horizontal direction, the momentum is conserved:
Initial momentum = Final momentum
p1 = p2
Step 5: Calculate the final velocity:
p1 = m2 * v2
2437.5 kg m/s = 300 kg * v2
v2 = 2437.5 kg m/s / 300 kg = 8.125 m/s
Step 6: Convert the final velocity back to km/h:
Final velocity (v2) = 8.125 m/s = 8.125 * (3600 s/h) / (1000 m/km) ≈ 29.25 km/h
Therefore, the speed of the trolley after the entire sandbag is empty is
approximately 29.25 km/h.
5.20 A body of mass 0.5 kg travels in a straight line with velocity v =a x3/2 where a = 5 m–1/2 s–1.What is the work done by the net force during its displacement from x = 0 to x = 2 m ?
Solution:
We can solve this problem using the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy.
* Find the initial and final velocities:
* Initial velocity (v1) at x = 0: v1 = a * 0^(3/2) = 0 m/s
* Final velocity (v2) at x = 2 m: v2 = a * 2^(3/2) = 5 * 2^(3/2) m/s
* Calculate the initial and final kinetic energies:
* Initial kinetic energy (KE1): KE1 = 0.5 * m * v1^2 = 0 J
* Final kinetic energy (KE2): KE2 = 0.5 * m * v2^2 = 0.5 * 0.5 kg * (5 * 2^(3/2) m/s)^2
* Calculate the work done:
* Work done (W) = Change in kinetic energy = KE2 - KE1
* W = 0.5 * 0.5 kg * (5 * 2^(3/2) m/s)^2 - 0 J
Simplifying the expression, we get:
W = 0.25 * 25 * 8 J = 50 J
Therefore, the work done by the net force during the displacement from x = 0 to x = 2 m is 50 Joules.
*
5.21 The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ? (b) What is the kinetic energy of the air ? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m^2, v = 36km/h and the density of air is 1.2 kg m–3. What is the electrical power produced ?
Solution:
(a) Mass of air passing through in time t:
The volume of air passing through the circle in time t is given by:
Volume = Area × Velocity × Time = A × v × t
The mass of this air is given by:
Mass = Density × Volume = ρ × A × v × t
Therefore, the mass of the air passing through the circle in time t is ρAvt.
(b) Kinetic energy of the air:
The kinetic energy of an object is given by:
Kinetic Energy = 1/2 × Mass × Velocity²
Substituting the mass of the air from part (a), we get:
Kinetic Energy = 1/2 × ρAvt × v² = 1/2 × ρAv³t
(c) Electrical power produced:
The windmill converts 25% of the wind's kinetic energy into electrical energy. Therefore, the electrical power produced is:
Electrical Power = 0.25 × Kinetic Energy per second
Substituting the kinetic energy per second from part (b) with t = 1 second, we get:
Electrical Power = 0.25 × 1/2 × ρAv³
Substituting the given values:
Electrical Power = 0.25 × 1/2 × 1.2 kg/m³ × 30 m² × (36 km/h × 1000 m/km / 3600 s/h)³
Calculating this expression gives:
Electrical Power ≈ 4500 W
Therefore, the electrical power produced by the windmill
is approximately 4500 watts.
5.22 A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force ? (b) Fat supplies 3.8 × 10^7J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Solution:
(a) To calculate the work done against gravity in one lift, we use the formula:
Work = Force × Distance
The force required to lift the mass is equal to its weight:
Force = Mass × Gravity = 10 kg × 9.81 m/s² = 98.1 N
So, the work done in one lift is:
Work_per_lift = 98.1 N × 0.5 m = 49.05 J
To find the total work done for 1000 lifts, we multiply the work per lift by the number of repetitions:
Total_work = 49.05 J/lift × 1000 lifts = 49050 J
Therefore, the dieter does 49,050 Joules of work against gravity.
(b) To calculate the mass of fat used, we first need to find the total energy required from fat. Since the efficiency of converting fat to mechanical energy is 20%, we need to divide the total work by the efficiency:
Energy_from_fat = Total_work / Efficiency = 49050 J / 0.2 = 245250 J
Now, we can calculate the mass of fat used by dividing the energy required from fat by the energy per kilogram of fat:
Mass_of_fat = Energy_from_fat / Energy_per_kg_fat = 245250 J / (3.8 × 10^7 J/kg) ≈ 0.00645 kg
Therefore, the dieter will use approximately 0.00645 kilograms of fat.
5.23 A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?
(b) Compare this area to that of the roof of a typical house.
Solution:
(a) To find the required area, we need to consider the effective power we can get from the solar panels.
Given:
* Power required = 8 kW = 8000 W
* Solar energy incident per square meter = 200 W/m²
* Efficiency of conversion = 20% = 0.2
So, the effective power per square meter = 200 W/m² * 0.2 = 40 W/m²
To get 8000 W, we need:
Area = Total Power / Effective Power per Square Meter
Area = 8000 W / 40 W/m²
Area = 200 m²
(b) Comparing this to a typical house roof:
Typical house roofs vary widely in size, but a common size for a single-family home might be around 100-200 square meters. So, the required solar panel area is comparable to the size of a typical house roof.
However, it's important to note that factors like roof orientation, shading, and local solar irradiance can affect the actual power output of solar panels. Additionally, the efficiency of solar panels can vary, so the required area might be slightly larger or smaller depending on the specific technology used.