Q: Q5. A body of mass 10 kg is initially at rest. A constant force of 20 N acts on it for 5 s. Calculate:\n(a) the work done by the force.\n(b) the change in kinetic energy of the body.

(a) The work done by the force is 1000 J.\n(b) The change in kinetic energy of the body is 1000 J.

Q: Q7. A block of mass 2 kg is pulled up a slope of length 4 m and inclination 30° with the horizontal. The force applied is 25 N parallel to the slope. The coefficient of kinetic frictio

(a) The work done by the applied force is 100 J.\n(b) The work done by the gravitational force is -40 J.\n(c) The work done by the frictional force is -13.86 J.\n(d) The net work done on the block is 46.14 J.\n(e) The change in kinetic energy of the block is 46.14 J.

Question 1

Q1. 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

Accepted Answer

The problem involves calculating work done by different forces and the change in kinetic energy, then verifying the Work-Energy Theorem. We first determine the forces acting on the body, calculate its acceleration and displacement, and then compute the required work and kinetic energy change.

**Given:**
*   Mass of the body, m = 2 kg
*   Initial velocity, u = 0 m/s (at rest)
*   Applied horizontal force, F_applied = 7 N
*   Coefficient of kinetic friction, μ_k = 0.1
*   Time, t = 10 s
*   Acceleration due to gravity, g = 9.8 m/s²

**Calculations:**

1.  **Normal force (N):** Since the body is on a horizontal table, the normal force balances the gravitational force.
    N = mg = 2 kg × 9.8 m/s² = 19.6 N

2.  **Frictional force (f_k):**
    f_k = μ_k × N = 0.1 × 19.6 N = 1.96 N

3.  **Net force (F_net):** The net force is the difference between the applied force and the frictional force.
    F_net = F_applied - f_k = 7 N - 1.96 N = 5.04 N

4.  **Acceleration (a):** Using Newton's second law (F_net = ma).
    a = F_net / m = 5.04 N / 2 kg = 2.52 m/s²

5.  **Displacement (s) in 10 s:** Using the kinematic equation s = ut + (1/2)at².
    s = (0 m/s × 10 s) + (1/2 × 2.52 m/s² × (10 s)²) = 0 + (1/2 × 2.52 × 100) m = 126 m

6.  **Final velocity (v) after 10 s:** Using the kinematic equation v = u + at.
    v = 0 m/s + (2.52 m/s² × 10 s) = 25.2 m/s

**(a) Work done by the applied force (W_applied):**
    W_applied = F_applied × s = 7 N × 126 m = 882 J

**(b) Work done by friction (W_friction):**
    Friction acts opposite to the direction of motion, so the work done by friction is negative.
    W_friction = -f_k × s = -1.96 N × 126 m = -246.96 J

**(c) Work done by the net force (W_net):**
    W_net = F_net × s = 5.04 N × 126 m = 635.04 J
    Alternatively, W_net = W_applied + W_friction = 882 J - 246.96 J = 635.04 J

**(d) Change in kinetic energy of the body (ΔKE):**
    Initial kinetic energy, KE_i = (1/2)mu² = (1/2) × 2 kg × (0 m/s)² = 0 J
    Final kinetic energy, KE_f = (1/2)mv² = (1/2) × 2 kg × (25.2 m/s)² = 1 kg × 635.04 m²/s² = 635.04 J
    ΔKE = KE_f - KE_i = 635.04 J - 0 J = 635.04 J

**Interpretation of results:**
We observe that the work done by the net force on the body (W_net = 635.04 J) is exactly equal to the change in its kinetic energy (ΔKE = 635.04 J). This result is consistent with the **Work-Energy Theorem**, which states that the net work done on an object is equal to the change in its kinetic energy.

Question 2

Q2. 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

Accepted Answer

The body of mass 2 kg moves under an applied force of 7 N on a table with kinetic friction. We first calculate the acceleration, then the displacement, and finally the work done by each force and the net force. The interpretation will be based on the Work-Energy Theorem.

(a) Work done by the applied force in 10 s is 700 J.
(b) Work done by friction in 10 s is -294 J.
(c) Work done by the net force on the body in 10 s is 406 J.
(d) Interpretation: The work done by the net force (406 J) is equal to the change in kinetic energy of the body (406 J), which is consistent with the Work-Energy Theorem.

Question 3

Q3. A body of mass 5 kg is acted upon by a force $\vec{F} = (-3\hat{i} + 6\hat{j})$ N. At time t = 0, the body is at rest at the origin. Compute the (a) position vector at t = 5 s, (b)

Accepted Answer

A body of mass 5 kg is subjected to a constant force. We will use Newton's second law to find acceleration, then kinematic equations to find the position vector, and finally the dot product of force and displacement to find the work done.

(a) The position vector at t = 5 s is $\vec{r} = (-7.5\hat{i} + 15\hat{j})$ m.
(b) The work done by the force in 5 s is 112.5 J.

Question 4

Q4. 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

Accepted Answer

The power delivered to the body at time t is proportional to t. Therefore, option (ii) is correct.

Question 5

Q5. A body of mass 10 kg is initially at rest. A constant force of 20 N acts on it for 5 s. Calculate:
(a) the work done by the force.
(b) the change in kinetic energy of the body.

Accepted Answer

(a) The work done by the force is 1000 J.
(b) The change in kinetic energy of the body is 1000 J.

Question 6

Q6. A body of mass 2 kg is thrown vertically upwards with an initial velocity of 20 m/s. What is its potential energy at the highest point? (Take g = 10 m/s²).

Accepted Answer

The potential energy of the body at the highest point is 400 J.

Question 7

Q7. A block of mass 2 kg is pulled up a slope of length 4 m and inclination 30° with the horizontal. The force applied is 25 N parallel to the slope. The coefficient of kinetic frictio

Accepted Answer

(a) The work done by the applied force is 100 J.
(b) The work done by the gravitational force is -40 J.
(c) The work done by the frictional force is -13.86 J.
(d) The net work done on the block is 46.14 J.
(e) The change in kinetic energy of the block is 46.14 J.

Question 8

Q8. A car of mass 1200 kg is moving with a speed of 54 km/h. What is the work done to stop the car?

Accepted Answer

The work done to stop the car is -135000 J or -135 kJ.

Question 9

Q9. A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to
(i) t^(1/2)
(ii) t^(3/4)
(iii) t^(3/2)\

Accepted Answer

The distance moved by the body in time t is proportional to t^(3/2). Therefore, option (iii) is correct.

Question 10

Q10. A body is constrained to move along the z-axis of a coordinate system. A constant force F given by F = -î + 2ĵ + 3k̂ N acts on the body. The work done by this force in moving the

Accepted Answer

The work done by this force in moving the body a distance of 4 m along the z-axis is 12 J. Therefore, option (ii) is correct.

Question 11

Q11. 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? Ob

Accepted Answer

The electron is faster than the proton. The ratio of the speed of the electron to the speed of the proton (v_e / v_p) is approximately 13.5.

Question 12

Q12. 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

Accepted Answer

The power delivered to the body at time t is proportional to t.

**Explanation:**
Given that the body is initially at rest, its initial velocity (u) = 0.
It undergoes one-dimensional motion with constant acceleration (a).

1.  **Velocity at time t:** Using the first equation of motion, v = u + at.
    Since u = 0, v = at.

2.  **Force acting on the body:** According to Newton's second law, F = ma.
    Since 'm' (mass) and 'a' (acceleration) are constant, the force F is constant.

3.  **Power delivered:** Power (P) is defined as the rate at which work is done, or the product of force and velocity in the direction of motion.
    P = F × v
    Substitute the expressions for F and v:
    P = (ma) × (at)
    P = m a^2 t

Since 'm' and 'a' are constants, m a^2 is also a constant. Therefore, Power P is directly proportional to time t.

P ∝ t

Thus, the correct option is (ii) t.

Question 13

Q13. A body is moving uniformly in a circle of radius R. Its speed is v. The work done by the centripetal force in moving it over half the circumference is
(i) Rv (ii) Rv^2/R (iii) zer

Accepted Answer

The work done by the centripetal force in moving the body over half the circumference is zero.

**Explanation:**
1.  **Nature of Centripetal Force:** In uniform circular motion, the centripetal force (F_c) always acts towards the center of the circular path. Its direction is always perpendicular to the instantaneous velocity vector of the body, which is tangential to the circle.
2.  **Definition of Work Done:** Work done (W) by a force is defined as the dot product of the force vector (F) and the displacement vector (s): W = F ⋅ s = F s cosθ, where θ is the angle between the force and displacement vectors.
3.  **Angle between Force and Displacement:** At any point in uniform circular motion, the centripetal force is directed radially inwards, while the instantaneous displacement (or velocity) is tangential to the circle. Therefore, the angle (θ) between the centripetal force and the instantaneous displacement is always 90°.
4.  **Work Done Calculation:** Since θ = 90°, cosθ = cos(90°) = 0.
    Therefore, W = F_c s cos(90°) = F_c s (0) = 0.

This holds true for any displacement, whether it's over half the circumference, a full circumference, or any small arc. The centripetal force does no work because it only changes the direction of the velocity, not its magnitude (speed).

Question 14

Q14. 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? Ob

Accepted Answer

The electron is faster than the proton. The ratio of the speed of the electron to the speed of the proton (v_e / v_p) is approximately 11.85.

Question 15

Q15. A body of mass 0.5 kg travels in a straight line with velocity v = ax^(3/2) where a = 5 m^(-1/2) s^(-1). What is the work done by the net force during its displacement from x = 0 t

Accepted Answer

The work done by the net force on the body can be calculated using the Work-Energy Theorem, which states that the net work done on an object is equal to the change in its kinetic energy.

Given:
Mass of the body, m = 0.5 kg
Velocity function, v = ax^(3/2)
Constant, a = 5 m^(-1/2) s^(-1)
Initial position, x_i = 0 m
Final position, x_f = 2 m

1.  **Calculate initial velocity (v_i) at x_i = 0 m:**
    v_i = a(0)^(3/2) = 0 m/s

2.  **Calculate initial kinetic energy (K_i):**
    K_i = (1/2)mv_i^2 = (1/2)(0.5 kg)(0 m/s)^2 = 0 J

3.  **Calculate final velocity (v_f) at x_f = 2 m:**
    v_f = a(2)^(3/2) = 5 m^(-1/2) s^(-1) * (2 m)^(3/2)
    v_f = 5 * (2 * √2) m/s = 10√2 m/s

4.  **Calculate final kinetic energy (K_f):**
    K_f = (1/2)mv_f^2 = (1/2)(0.5 kg)(10√2 m/s)^2
    K_f = (1/2)(0.5)(100 * 2) J
    K_f = (1/2)(0.5)(200) J
    K_f = 0.5 * 100 J = 50 J

5.  **Calculate the work done by the net force (W_net) using the Work-Energy Theorem:**
    W_net = ΔK = K_f - K_i
    W_net = 50 J - 0 J = 50 J

Therefore, the work done by the net force during the displacement from x = 0 to x = 2 m is 50 J.

Question 16

Q16. 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? Ob

Accepted Answer

To determine which particle is faster and to find the ratio of their speeds, we will use the formula for kinetic energy, K = (1/2)mv^2, and solve for velocity, v = √(2K/m).

Given:
Kinetic energy of electron, K_e = 10 keV
Kinetic energy of proton, K_p = 100 keV
Mass of electron, m_e = 9.11 × 10^(-31) kg
Mass of proton, m_p = 1.67 × 10^(-27) kg
Conversion factor, 1 eV = 1.60 × 10^(-19) J

**Step 1: Convert kinetic energies from keV to Joules.**
K_e = 10 keV = 10 × 10^3 eV = 10^4 eV
K_e = 10^4 × (1.60 × 10^(-19) J) = 1.60 × 10^(-15) J

K_p = 100 keV = 100 × 10^3 eV = 10^5 eV
K_p = 10^5 × (1.60 × 10^(-19) J) = 1.60 × 10^(-14) J

**Step 2: Calculate the speed of the electron (v_e).**
v_e = √(2K_e / m_e)
v_e = √(2 × 1.60 × 10^(-15) J / 9.11 × 10^(-31) kg)
v_e = √(3.20 × 10^(-15) / 9.11 × 10^(-31)) m/s
v_e = √(0.35126 × 10^(16)) m/s
v_e = √(3.5126 × 10^(15)) m/s
v_e ≈ 5.927 × 10^7 m/s

**Step 3: Calculate the speed of the proton (v_p).**
v_p = √(2K_p / m_p)
v_p = √(2 × 1.60 × 10^(-14) J / 1.67 × 10^(-27) kg)
v_p = √(3.20 × 10^(-14) / 1.67 × 10^(-27)) m/s
v_p = √(1.916 × 10^(13)) m/s
v_p ≈ 4.377 × 10^6 m/s

**Step 4: Compare the speeds.**
v_e ≈ 5.927 × 10^7 m/s
v_p ≈ 4.377 × 10^6 m/s
Since 5.927 × 10^7 m/s > 4.377 × 10^6 m/s, the electron is faster than the proton.

**Step 5: Obtain the ratio of their speeds (v_e / v_p).**
Ratio = v_e / v_p = (5.927 × 10^7 m/s) / (4.377 × 10^6 m/s)
Ratio ≈ 13.54

Thus, the electron is faster than the proton, and the ratio of the electron's speed to the proton's speed is approximately 13.54.

Question 17

Q17. 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 he

Accepted Answer

Let's break down this problem into parts, calculating the work done by gravity and then by the resistive force.

Given:
Radius of raindrop, r = 2 mm = 2 × 10^(-3) m
Total height, H = 500 m
Acceleration due to gravity, g = 10 m s^(-2)
Density of water, ρ = 10^3 kg m^(-3) (standard value)
Final speed on reaching the ground, v_f = 10 m s^(-1)
Initial speed, v_i = 0 m s^(-1) (starts from rest)

**Part 1: Calculate the mass of the raindrop.**
Volume of raindrop, V = (4/3)πr^3
V = (4/3) × 3.14159 × (2 × 10^(-3) m)^3
V = (4/3) × 3.14159 × 8 × 10^(-9) m^3
V ≈ 3.351 × 10^(-8) m^3

Mass of raindrop, m = ρ × V
m = 10^3 kg m^(-3) × 3.351 × 10^(-8) m^3
m = 3.351 × 10^(-5) kg

**Part 2: Work done by gravitational force in the first half of the journey.**
Height for the first half, h_1 = H/2 = 500 m / 2 = 250 m
Work done by gravity, W_g1 = mgh_1
W_g1 = (3.351 × 10^(-5) kg) × (10 m s^(-2)) × (250 m)
W_g1 = 3.351 × 10^(-4) × 250 J
W_g1 = 0.0083775 J ≈ 8.38 × 10^(-3) J

**Part 3: Work done by gravitational force in the second half of the journey.**
Height for the second half, h_2 = H/2 = 250 m
Work done by gravity, W_g2 = mgh_2
W_g2 = (3.351 × 10^(-5) kg) × (10 m s^(-2)) × (250 m)
W_g2 = 0.0083775 J ≈ 8.38 × 10^(-3) J

**Part 4: Work done by the resistive force in the entire journey.**
We use the Work-Energy Theorem for the entire journey:
W_net = ΔK = K_f - K_i

The net work done (W_net) is the sum of work done by gravity (W_g_total) and work done by the resistive force (W_r).
W_net = W_g_total + W_r

Total work done by gravity, W_g_total = mgH
W_g_total = (3.351 × 10^(-5) kg) × (10 m s^(-2)) × (500 m)
W_g_total = 3.351 × 10^(-4) × 500 J
W_g_total = 0.016755 J ≈ 0.0168 J

Initial kinetic energy, K_i = (1/2)mv_i^2 = (1/2)m(0)^2 = 0 J
Final kinetic energy, K_f = (1/2)mv_f^2
K_f = (1/2) × (3.351 × 10^(-5) kg) × (10 m s^(-1))^2
K_f = (1/2) × (3.351 × 10^(-5)) × 100 J
K_f = 3.351 × 10^(-5) × 50 J
K_f = 1.6755 × 10^(-3) J ≈ 0.00168 J

Now, apply the Work-Energy Theorem:
W_g_total + W_r = K_f - K_i
0.016755 J + W_r = 0.0016755 J - 0 J
W_r = 0.0016755 J - 0.016755 J
W_r = -0.0150795 J ≈ -0.0151 J

The negative sign indicates that the resistive force acts opposite to the direction of motion, as expected.

**Summary of Results:**
*   Work done by gravitational force in the first half = 8.38 × 10^(-3) J
*   Work done by gravitational force in the second half = 8.38 × 10^(-3) J
*   Work done by the resistive force in the entire journey = -0.0151 J

Question 18

Q18. A body of mass 0.5 kg travels in a straight line with velocity v = ax^(3/2) where a = 5 m^(-1/2) s^(-1). What is the work done by the net force during its displacement from x = 0 t

Accepted Answer

The work done by the net force on the body during its displacement from x = 0 to x = 2 m is 50 J.

Question 19

Q19. A block of mass 1 kg is pushed up a surface inclined to the horizontal at an angle of 30° by a force of 10 N parallel to the inclined surface. The coefficient of kinetic friction b

Accepted Answer

The acceleration of the block up the incline is approximately 3.268 m/s². However, the distance the block will travel cannot be determined without additional information such as the time for which the force is applied or the final velocity of the block.

Question 20

Q20. 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

Accepted Answer

The power delivered to the body at time t is proportional to t. Therefore, option (ii) is correct.

Question 21

Q21. A body is moving uniformly in a circle of radius R. Its speed is v. The magnitude of change in its velocity in time t is
(i) zero
(ii) v^2/R * t
(iii) v * t
(iv) 2v sin(vt/2R)

Accepted Answer

The magnitude of change in its velocity in time t is 2v sin(vt/2R). Therefore, option (iv) is correct.

Question 22

Q22. 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? Ob

Accepted Answer

The electron is faster than the proton. The ratio of their speeds (v_e / v_p) is approximately 11.8.

Question 23

Q23. A body of mass 0.5 kg travels in a straight line with velocity v = ax^(3/2) where a = 5 m^(-1/2) s^(-1). What is the work done by the net force during its displacement from x = 0 t

Accepted Answer

The work done by the net force during the displacement from x = 0 to x = 2 m is 50 J.

Question 24

Q24. A block of mass 1 kg is pushed up a surface inclined at an angle of 30° with the horizontal by a force of 10 N parallel to the inclined surface (Fig. 6.17). The coefficient of kine

Accepted Answer

Given:
Mass of block, m = 1 kg
Angle of inclination, θ = 30°
Applied force, F_app = 10 N (parallel to incline, upwards)
Coefficient of kinetic friction, μ_k = 0.2
Displacement, d = 10 m (up the incline)
Acceleration due to gravity, g = 10 m s^(-2)

(a) Work done against gravity = 50 J
(b) Work done against the frictional force = 17.32 J
(c) Work done by the applied force = 100 J
(d) Increase in kinetic energy = 32.68 J

Question 25

Q25. A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m^3 in 15 min. If the tank is 40 m above the ground and the efficiency of the pump is 30%, ho

Accepted Answer

The electric power consumed by the pump is approximately 43555.56 W or 43.56 kW.

Question 26

Q26. Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another identical ball bearing moving with a speed V. If the collision

Accepted Answer

The correct option is (a). One ball bearing moves with speed V, the other two remain at rest.

Question 27

Q27. A body of mass 0.5 kg travels in a straight line with velocity v = ax^(3/2) where a = 5 m^(-1/2) s^(-1). What is the work done by the net force during its displacement from x = 0 t

Accepted Answer

The work done by the net force on the body during its displacement from x = 0 to x = 2 m is 50 J.

Question 28

Q28. 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? (E

Accepted Answer

The electron is faster than the proton.

Question 29

Q29. 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

Accepted Answer

The power delivered to the body at time t is proportional to t.

**Explanation:**
Given:
Initial velocity, u = 0 (body is initially at rest)
Acceleration, a = constant

1.  **Velocity at time t:**
    Using the first equation of motion, v = u + at
    Since u = 0, v = at

2.  **Force acting on the body:**
    According to Newton's second law, F = ma
    Since 'a' is constant, 'F' is also constant.

3.  **Power delivered:**
    Power (P) is defined as the rate at which work is done, or the product of force and velocity.
    P = F ⋅ v
    Substitute F = ma and v = at:
    P = (ma) ⋅ (at)
    P = m a² t

Since 'm' (mass) and 'a' (constant acceleration) are constants, m a² is also a constant. Therefore, Power P is directly proportional to time t.

P ∝ t

Thus, the correct option is (ii) t.

Question 30

Q30. A body is moving uniformly in a circle of radius R. Its speed is v. The work done by the centripetal force in moving it over half the circumference is
(i) Rv
(ii) Rv^2
(iii) Rv^

Accepted Answer

The work done by the centripetal force in moving the body over half the circumference is 0.

**Explanation:**
1.  **Centripetal Force Direction:** When a body moves uniformly in a circle, the centripetal force always acts towards the center of the circle, along the radius. Its direction is perpendicular to the instantaneous velocity of the body. 2.  **Displacement Direction:** The instantaneous displacement of the body is always tangential to the circular path, in the direction of its velocity. 3.  **Angle between Force and Displacement:** Since the centripetal force acts radially inwards and the instantaneous displacement is tangential, the angle (θ) between the centripetal force and the instantaneous displacement is always 90 degrees (θ = 90°). 4.  **Work Done Formula:** Work done (W) by a force (F) over a displacement (d) is given by W = F ⋅ d = F d cos(θ). 5.  **Calculation:** Substituting θ = 90° into the work done formula:
    W = F d cos(90°)
    Since cos(90°) = 0,
    W = F d (0)
    W = 0

Therefore, the work done by the centripetal force is always zero, regardless of the distance covered (half circumference, full circumference, or any part) as long as the motion is circular and the force is purely centripetal.

Chapter 6: Work, Energy and Power