🔹 Topic: MCQs on Work done by Variable Force | MCQs: 40 🔹
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1. A spring of force constant 10 N/m has an initial stretch 0.20 m. In changing the stretch to 0.25 m, the increase in potential energy is about
Correct answer: (a) 0.1 joule
\( U = \frac{1}{2}k(x_f^2 - x_i^2) = \frac{1}{2} \times 10 \times (0.25^2 - 0.20^2) = 5 \times (0.0625 - 0.04) = 0.1125 \approx 0.1 \) J
2. Which one of the following is not a conservative force
Correct answer: (d) Frictional force
Frictional force is non-conservative because work done depends on path length.
3. A cord is used to lower vertically a block of mass M by a distance d with constant downward acceleration g/4. Work done by the cord on the block is
Correct answer: (c) \( -\frac{3}{4}Mgd \)
Tension \( T = M(g - a) = M(g - g/4) = \frac{3}{4}Mg \)
Work \( W = \vec{T} \cdot \vec{d} = -Td = -\frac{3}{4}Mgd \)
Work \( W = \vec{T} \cdot \vec{d} = -Td = -\frac{3}{4}Mgd \)
4. Two springs have their force constant as \( k_1 \) and \( k_2 \) (\( k_1 > k_2 \)). When they are stretched by the same force
Correct answer: (c) More work is done in case of second spring
\( W = \frac{F^2}{2k} \), so softer spring (smaller k) requires more work.
5. The force constant of a wire is \( k \) and that of another wire is \( 2k \). When both the wires are stretched through same distance, then the work done
Correct answer: (b) \( W_2 = 2W_1 \)
\( W = \frac{1}{2}kx^2 \), so \( \frac{W_2}{W_1} = \frac{2k}{k} = 2 \)
6. A particle moves under the effect of a force \( F = Cx \) from \( x = 0 \) to \( x = x_0 \). The work done in the process is
Correct answer: (b) \( \frac{1}{2}Cx_0^2 \)
\( W = \int_0^{x_0} Cx dx = \frac{C}{2}x_0^2 \)
7. A position dependent force \( F = (7 - 2x + 3x^2) \) N acts on a small body of mass 2 kg and displaces it from \( x = 0 \) to \( x = 5 \) m. The work done in joules is
Correct answer: (d) 135
\( W = \int_0^5 (7 - 2x + 3x^2) dx = [7x - x^2 + x^3]_0^5 = 135 \) J
8. The potential energy of a certain spring when stretched through a distance 'S' is 10 joule. The amount of work (in joule) that must be done on this spring to stretch it through an additional distance 'S' will be
Correct answer: (a) 30
PE ∝ stretch², so 2S stretch gives 4×10 = 40 J. Additional work = 40 - 10 = 30 J
9. A body of mass 0.1 kg moving with a velocity of 10 m/s hits a spring (fixed at the other end) of force constant 1000 N/m and comes to rest after compressing the spring. The compression of the spring is
Correct answer: (a) 0.1 m
\( \frac{1}{2}mv^2 = \frac{1}{2}kx^2 \) ⇒ \( x = v\sqrt{m/k} = 10\sqrt{0.1/1000} = 0.1 \) m
10. A body of mass 3 kg is under a force, which causes a displacement in it is given by \( s = t^3/3 \) (in m). Find the work done by the force in first 2 seconds
Correct answer: (d) 24 J
\( v = ds/dt = t^2 \), \( a = dv/dt = 2t \)
\( F = ma = 6t \), \( W = \int F ds = \int_0^2 6t \cdot t^2 dt = \int_0^2 6t^3 dt = 24 \) J
\( F = ma = 6t \), \( W = \int F ds = \int_0^2 6t \cdot t^2 dt = \int_0^2 6t^3 dt = 24 \) J
11. A spring when stretched by 2 mm its potential energy becomes 4 J. If it is stretched by 10 mm, its potential energy is equal to
Correct answer: (d) None
\( U \propto x^2 \), so \( \frac{U_2}{4} = \left(\frac{10}{2}\right)^2 = 25 \) ⇒ \( U_2 = 100 \) J
12. When a 1.0 kg mass hangs attached to a spring of length 50 cm, the spring stretches by 2 cm. The mass is pulled down until the length of the spring becomes 60 cm. What is the amount of elastic energy stored in the spring in this condition, if g = 10 m/s²
Correct answer: (c) 2.5 Joule
\( k = mg/x = (1 \times 10)/0.02 = 500 \) N/m
Total stretch = 10 cm = 0.1 m
\( U = \frac{1}{2}kx^2 = \frac{1}{2} \times 500 \times (0.1)^2 = 2.5 \) J
Total stretch = 10 cm = 0.1 m
\( U = \frac{1}{2}kx^2 = \frac{1}{2} \times 500 \times (0.1)^2 = 2.5 \) J
13. A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is
Correct answer: (b) 8 J
\( W = \frac{1}{2}k(x_2^2 - x_1^2) = \frac{1}{2} \times 800 \times (0.15^2 - 0.05^2) = 8 \) J
14. A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k = 50 N/m. The maximum compression of the spring would be
Correct answer: (a) 0.15 m
\( \frac{1}{2}mv^2 = \frac{1}{2}kx^2 \) ⇒ \( x = v\sqrt{m/k} = 1.5\sqrt{0.5/50} = 0.15 \) m
15. When a spring is stretched by 2 cm, it stores 100 J of energy. If it is stretched further by 2 cm, the stored energy will be increased by
Correct answer: (c) 300 J
\( U \propto x^2 \), so 4 cm stretch gives 4×100 = 400 J. Additional energy = 400 - 100 = 300 J
16. A spring of spring constant 5 × 10³ N/m is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is
Correct answer: (c) 18.75 N-m
\( W = \frac{1}{2}k(x_2^2 - x_1^2) = \frac{1}{2} \times 5000 \times (0.1^2 - 0.05^2) = 18.75 \) J
17. Two springs of spring constants 1500 N/m and 3000 N/m respectively are stretched with the same force. They will have potential energy in the ratio
Correct answer: (c) 2 : 1
\( U = \frac{F^2}{2k} \), so \( \frac{U_1}{U_2} = \frac{k_2}{k_1} = \frac{3000}{1500} = 2 \)
18. A spring 40 mm long is stretched by the application of a force. If 10 N force required to stretch the spring through 1 mm, then work done in stretching the spring through 40 mm is
Correct answer: (d) 8 J
\( k = F/x = 10/0.001 = 10000 \) N/m
\( W = \frac{1}{2}kx^2 = \frac{1}{2} \times 10000 \times (0.04)^2 = 8 \) J
\( W = \frac{1}{2}kx^2 = \frac{1}{2} \times 10000 \times (0.04)^2 = 8 \) J
19. A spring with spring constant k is extended from \( x_1 \) to \( x_2 \). The work done will be
Correct answer: (b) \( \frac{1}{2}k(x_2^2 - x_1^2) \)
Work done in stretching spring: \( W = \frac{1}{2}k(x_2^2 - x_1^2) \)
20. The potential energy of a body is given by, \( U = A - Bx^2 \) (Where x is the displacement). The magnitude of force acting on the particle is
Correct answer: (b) Proportional to x
\( F = -\frac{dU}{dx} = 2Bx \), so force is proportional to x
21. If a long spring is stretched by 0.02 m, its potential energy is U. If the spring is stretched by 0.1 m, then its potential energy will be
Correct answer: (d) 25U
\( U \propto x^2 \), so \( \frac{U_2}{U} = \left(\frac{0.1}{0.02}\right)^2 = 25 \)
22. Natural length of a spring is 60 cm, and its spring constant is 4000 N/m. A mass of 20 kg is hung from it. The extension produced in the spring is, (Take g = 9.8 m/s²)
Correct answer: (a) 4.9 cm
\( x = \frac{mg}{k} = \frac{20 \times 9.8}{4000} = 0.049 \) m = 4.9 cm
23. The spring extends by x on loading, then energy stored by the spring is : (if T is the tension in spring and k is spring constant)
Correct answer: (a) \( \frac{T^2}{2k} \)
\( U = \frac{1}{2}kx^2 \) and \( T = kx \), so \( U = \frac{T^2}{2k} \)
24. The potential energy between two atoms in a molecule is given by \( U(x) = \frac{a}{x^{12}} - \frac{b}{x^6} \); where a and b are positive constants and x is the distance between the atoms. The atom is in stable equilibrium when
Correct answer: (a) \( x = \left(\frac{2a}{b}\right)^{1/6} \)
For equilibrium: \( \frac{dU}{dx} = 0 \) ⇒ \( x = \left(\frac{2a}{b}\right)^{1/6} \)
25. A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to
Correct answer: (b) \( x^2 \)
Retardation \( a = -kx \), work done = loss in KE = \( \int F dx = \int -mkx dx \propto x^2 \)
26. A spring with spring constant k when stretched through 1 cm, the potential energy is U. If it is stretched by 4 cm. The potential energy will be
Correct answer: (c) 16U
\( U \propto x^2 \), so \( \frac{U_2}{U} = \left(\frac{4}{1}\right)^2 = 16 \)
Work Done by Variable Force (Additional Questions)
27. A force \( F = 3x^2 + 2x \) (in Newtons) acts on a particle. The work done by this force as the particle moves from x = 1 m to x = 3 m is:
Correct answer: (d) 34 J
\( W = \int_1^3 (3x^2 + 2x) dx = [x^3 + x^2]_1^3 = 34 \) J
28. A particle is acted upon by a force \( F = -kx^3 \) where k is a positive constant. The work done by this force when the particle moves from x = a to x = 0 is:
Correct answer: (a) \( \frac{ka^4}{4} \)
\( W = \int_a^0 (-kx^3) dx = \frac{ka^4}{4} \)
29. A variable force F acting on a particle varies with position x as \( F = 5e^{-2x} \) N. The work done by this force as the particle moves from x = 0 to x = ∞ is:
Correct answer: (a) 2.5 J
\( W = \int_0^∞ 5e^{-2x} dx = \frac{5}{2} = 2.5 \) J
30. A force \( F = 2x + 3y \) N acts on a particle. The work done by this force as the particle moves from (0,0) to (2,4) along the path y = x² is:
Correct answer: (d) 28 J
\( W = \int_0^2 (2x + 3x^2 + x^2 \cdot 2x) dx = \int_0^2 (2x + 3x^2 + 2x^3) dx = 28 \) J
31. A particle moves along the x-axis under the action of a force \( F = -ax + bx^3 \) where a and b are positive constants. The work done by this force when the particle moves from x = 0 to x = \( \sqrt{a/b} \) is:
Correct answer: (a) \( -\frac{a^2}{4b} \)
\( W = \int_0^{\sqrt{a/b}} (-ax + bx^3) dx = -\frac{a^2}{4b} \)
32. A particle moves along the curve \( y = x^2 \) from (0,0) to (1,1) under the action of a force \( \vec{F} = y\hat{i} + x\hat{j} \). The work done by the force is:
Correct answer: (b) 2/3 J
\( W = \int_0^1 (x^2 dx + x \cdot 2x dx) = \int_0^1 (x^2 + 2x^2) dx = \frac{2}{3} \) J
33. A force \( F = 10/x^2 \) N acts on a particle. The work done by this force as the particle moves from x = 1 m to x = 10 m is:
Correct answer: (a) 9 J
\( W = \int_1^{10} \frac{10}{x^2} dx = 9 \) J
34. A force \( \vec{F} = (2xy + z)\hat{i} + x^2\hat{j} + x\hat{k} \) N acts on a particle. The work done by this force as the particle moves from (0,0,0) to (1,1,1) along the path x = t, y = t², z = t³ is:
Correct answer: (d) 2 J
\( W = \int_0^1 (2t^3 + t^3 + t^2 \cdot 2t + t \cdot 3t^2) dt = 2 \) J
35. A force \( F = 5 \sin(\pi x/2) \) N acts on a particle constrained to move along the x-axis. The work done by this force as the particle moves from x = 0 to x = 1 m is:
Correct answer: (b) \( \frac{10}{\pi} \) J
\( W = \int_0^1 5 \sin(\frac{\pi x}{2}) dx = \frac{10}{\pi} \) J
36. A force \( F = kx^2 \) acts on a particle, where k is a constant and x is the displacement from equilibrium. The work done by this force as the particle moves from x = a to x = 2a is:
Correct answer: (a) \( \frac{7ka^3}{3} \)
\( W = \int_a^{2a} kx^2 dx = \frac{7ka^3}{3} \)
37. A particle moves along the x-axis under a force that varies with position as \( F = -kx + ax^3 \), where k and a are positive constants. The work done by this force as the particle moves from x = 0 to x = \( \sqrt{k/a} \) is:
Correct answer: (a) \( -\frac{k^2}{4a} \)
\( W = \int_0^{\sqrt{k/a}} (-kx + ax^3) dx = -\frac{k^2}{4a} \)
38. A force \( \vec{F} = (2xy + z^3)\hat{i} + x^2\hat{j} + 3xz^2\hat{k} \) N acts on a particle. The work done by this force as the particle moves from (0,0,0) to (1,1,1) along the path x = t, y = t², z = t³ is:
Correct answer: (d) 2 J
\( W = \int_0^1 (2t^3 + t^6 + t^2 \cdot 2t + 3t^4 \cdot 3t^2) dt = 2 \) J
39. A force \( F = 5x^2 \) N acts on a particle constrained to move along the x-axis. The work done by this force as the particle moves from x = 1 m to x = 2 m is:
Correct answer: (a) \( \frac{35}{3} \) J
\( W = \int_1^2 5x^2 dx = \frac{35}{3} \) J
40. A particle moves along the x-axis under a force \( F = 3x^2 - 2x \). The work done by this force as the particle moves from x = -1 to x = 2 is:
Correct answer: (b) 4.5 J
\( W = \int_{-1}^2 (3x^2 - 2x) dx = [x^3 - x^2]_{-1}^2 = 4.5 \) J
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