Physics Questions and Answers – Work Done by a Variable Force

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This set of Physics Aptitude Test for NEET Exam focuses on “Work Done by a Variable Force”.

1. A particle of mass 0.1 kg is subjected to a force which varies with distance as shown in the figure. What is the work done?

a) 20 J
b) 40 J
c) 60 J
d) 80 J
View Answer

Answer: d
Explanation: The work done (W) is given by;
W = Force x Displacement
The area enclosed by the curves in the graph shown gives the work done;
Enclosed area = [(1/2) x 4 x 10] + (4 x 10) + [(1/2) x 4 x 10]
= 80 Nm
= 80 J.
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2. The length of a smooth inclined plane of 30-degree inclination is 5 m. The work done in moving a 10 kg mass from the bottom of the inclined plane to the top is _____ Joules. (Assume g = 10m/s2)
a) 250
b) 1000
c) 1250
d) 500
View Answer

Answer: a
Explanation: Length (l) = 5 m
Height = Displacement = 5 x sin30
= 5 x (1/2)
= 2.5
Force on object = m x g
= 10 X 10
= 100 N
Work = Force x Displacement
= 100 x 2.5
= 250 Joules.

3. A rope has a uniform mass density of 0.4 kg/m. The rope is 10 m long and is lowered into a 10 m deep pit. The bottom part of the rope just touches the bottom of the pit. What is the work done to pull the rope out of the well completely? (Assume g = 10m/s2)
a) 200 J
b) 400 J
c) 2000 J
d) 4000 J
View Answer

Answer:a
Explanation: Mass density (p) = 0.4 kg/m
Length (l) = 10 m
Force = f(x)
= m(x)*g
= 0.4x*g
= 4x
Displacement (d) = 10 m
Work = \(\int_{0}^{l}\)f(x)dx
= \(\int_{0}^{10}\)4xdx
= 200 J.
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4. A bucket filled with water weighing 20 kg is raised from a well of depth 20 m. If the linear density of the rope is 0.2 kg/m, the work done is _____ (Assume g = 10m/s2)
a) 4000 J
b) 4040 J
c) 4400 J
d) 4800 J
View Answer

Answer: c
Explanation: Mass density (p) = 0.2 kg/m
Length (l) = 20 m
Force = f(x)
= m(x)*g
= 0.2x*g
= 2x
Displacement (d) = 20 m
Work = \(\int_{0}^{l}\)f(x)dx+ (20*g*20)
= \(\int_{0}^{20}\)2xdx + 4000
= 4400 J.

5. A force F = (5i – 3j +2k) N moves a particle from r1 = (2i + 7j + 4k) m to r2 = (5i + 2j + 8k) m. What is the work done by the force?
a) 18 J
b) 28 J
c) 38 J
d) 48 J
View Answer

Answer: c
Explanation: Displacement (d) = r2 – r1
= (3i – 5j + 4k) m
Work = Force . Displacement; [Dot product of vectors]
= (5i – 3j +2k) .(3i – 5j + 4k)
= 38 J.
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6. A person of mass 50 kg carrying a load of 20 kg walks up a staircase. The width and height of each step are 0.25 m and 0.2 m respectively. What is the work done by the man in walking 20 steps?(Assume g = 10m/s2)
a) 800 J
b) 1600 J
c) 2000 J
d) 2800 J
View Answer

Answer: d
Explanation: Total mass (m) = 50 + 20 = 70 kg
Vertical displacement = 0.2 x 20 = 4m
Horizontal displacement = 0.25 x 20 = 5m
Displacement = (5i + 4j) m
Force = m x acceleration
= 70 x (0i + 10j)
= (700j) N
Work = Force .Displacement
= (0i + 700j) . (5i + 4j)
= 2800 J.

7. A chain of length 4 m is kept on a table. 2 m of the chain hangs freely from the table’s edge. The mass of the whole chain is 8 kg. What is the work done to pull the whole chain onto the table? (Assume g = 10m/s2)
a) 20 J
b) 40 J
c) 60 J
d) 80 J
View Answer

Answer: b
Explanation: Mass per length of chain (p) = 8/4 = 2 kg/m
Mass of ‘x’ m of length = 2x kg
Force = \(\int_{0}^{2}\)(2x * g)dx
= \(\int_{0}^{2}\)20xdx
= 40 J.
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8. The area under the force (F) versus displacement (x) graph gives work.
a) True
b) False
View Answer

Answer: a
Explanation: The area under this graph would be \(\int_{x1}^{x2}\)F * dx, which is work. Here x1 & x2 are two positions on the x-axis. Hence the statement is true.

9. An object of mass 1kg is subjected to a variable force in x-direction. The force function is F = x2 N. What is the work done in moving the object from x = 2 to 5?
a) 40
b) 39
c) 38
d) 69
View Answer

Answer: b
Explanation: The work is given by formula W = \(\int_{x1}^{x2}\)F * dx. So, W = \(\int_{2}^{5}\)x2 * dx = (53-23)/3 = 39 J.
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10. A particle is acted upon by a force in x direction which varies as shown in the graph. What is the work done on the particle?

a) 100 J
b) 10 J
c) 200 J
d) 150 J
View Answer

Answer: b
Explanation: The work done is the area under the force versus displacement graph. The area can be calculated as area of trapezium i.e. 0.5*(sum of parallel sides)*height = 0.5*(6+4)*2 = 10.

11. What is the SI unit of variable force?
a) Newtons
b) Joules
c) Ampere
d) Mole
View Answer

Answer: a
Explanation: The SI unit of force is Newtons. While Joules, Ampere, and Mole are SI units of energy, current and amount of substance.

12. A spring is stretched by a length x, then the potential energy stored in it changes by 0.5*k*x2, where k is the spring constant.
a) True
b) False
View Answer

Answer: a
Explanation: The potential energy of a stretched spring is given by 0.5*k*x2 where x is the elongation in the spring.

13. A force of 600 N elongates a spring from its natural length of 18 cm to a length of 20 cm. What is the work done?
a) 116J
b) 114J
c) 118J
d) 120J
View Answer

Answer: b
Explanation: The force on a spring stretched by x meters is given by F = k*x where k is the spring constant. So, 600 = k*(2/100). Now k = 3*104, and hence F = 30000*x. W = \(\int_{x1}^{x2}\)F * dx = \(\int_{0.18}^{0.2}\)30000 * x * dx = 114 J.

14. Which of the following is not a variable force?
a) F = x
b) F = y2
c) F = 2
d) F = 2*t
View Answer

Answer: c
Explanation: F = x, F = x2 varies with x and y co-ordinate respectively. F = 2*t varies with time but F = 2 is always constant.

15. Which of the following is not a conservative force?
a) Gravitational force
b) Spring force
c) Friction force
d) Magnetic force
View Answer

Answer: c
Explanation: Friction force causes energy losses in form of heat so it is a non-conservative force while gravitational, spring and magnetic forces are all conservative.

16. A force F = x + 1 N acts on a particle in x direction. What is the work done from x = 2m to 5 m?
a) 10.5 J
b) 14.5 J
c) 13.5 J
d) 0.5 J
View Answer

Answer: c
Explanation: Work done is given by W = \(\int_{x1}^{x2}\)F * dx. So, W = \(\int_{2}^{5}\)(x + 1) * dx = [(52/2 + 5) – (22/2+2)] = 13.5.

17. What is the value of spring constant if force of 200 N is acting on it and the spring is compressed by 2m?
a) 100 kgs-2
b) 100 kgs-1
c) 100 kg2s2
d) 100 Kgs
View Answer

Answer: a
Explanation: Force on a elongated/compressed spring is given by F = k*x where k is spring constant and x is elongation/compression. So, 200 = k*2 which gives k = 100 kgs-2.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter