Unit 4: Work Energy and Power — Online MCQ Test
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Practice Unit 4: Work Energy and Power with a free chapter-wise online MCQ test.
This chapter covers: Work done by a constant force and a variable force; kinetic and potential energies work-energy theorem power. The potential energy of spring conservation of mechanical energy conse....
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Unit 4: Work Energy and Power — Important Questions & Answers
Light with an average flux of 20 W/cm² falls on a non-reflecting surface at normal incidence having surface area 20 cm². The energy received by the surface during a time span of 1 minute is:
- A. 12 x 10³ J
- B. 24 x 10³ J
- C. 48 x 10³ J
- D. 10 x 10³ J
Answer: B. 24 x 10³ J
Power = Intensity * Area = 20 W/cm² * 20 cm² = 400 W. Energy = Power * Time = 400 W * 60 s = 24,000 J = 24 x 10³ J.
Power = Intensity * Area = 20 W/cm² * 20 cm² = 400 W. Energy = Power * Time = 400 W * 60 s = 24,000 J = 24 x 10³ J.
The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m. If FA and FB are the forces applied by the breaks on cars A and B, respectively, then the ratio FA/FB is
- A. 3/2
- B. 2/3
- C. 1/3
- D. 1/2
Answer: B. 2/3
Work done = Force * distance = Change in KE. FA * 1000 = 100 => FA = 0.1. FB * 1500 = 225 => FB = 0.15. FA/FB = 0.1 / 0.15 = 2/3.
Work done = Force * distance = Change in KE. FA * 1000 = 100 => FA = 0.1. FB * 1500 = 225 => FB = 0.15. FA/FB = 0.1 / 0.15 = 2/3.
The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m. If FA and FB are the forces applied by the breaks on cars A and B, respectively, then the ratio FA/FB is
- A. 3/2
- B. 2/3
- C. 1/3
- D. 1/2
Answer: B. 2/3
Work done = Force × displacement = Change in Kinetic Energy. FA * 1000 = 100 and FB * 1500 = 225. FA = 0.1, FB = 0.15. FA/FB = 0.1/0.15 = 10/15 = 2/3.
Work done = Force × displacement = Change in Kinetic Energy. FA * 1000 = 100 and FB * 1500 = 225. FA = 0.1, FB = 0.15. FA/FB = 0.1/0.15 = 10/15 = 2/3.
A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take g = 9.8 m/s^2)
- A. 21 Ns
- B. 7 Ns
- C. 0
- D. 84 Ns
Answer: A. 21 Ns
Initial velocity v1 = sqrt(2gh) = sqrt(2*9.8*40) = 28 m/s. Final velocity v2 = sqrt(2*9.8*10) = 14 m/s. Impulse = m(v1 + v2) = 0.5*(28+14) = 0.5*42 = 21 Ns.
Initial velocity v1 = sqrt(2gh) = sqrt(2*9.8*40) = 28 m/s. Final velocity v2 = sqrt(2*9.8*10) = 14 m/s. Impulse = m(v1 + v2) = 0.5*(28+14) = 0.5*42 = 21 Ns.
A body of mass 1 kg begins to move under the action of a time dependent force F = (3t^2 i_hat + 2t j_hat)N. What power will be developed by the force at the time t?
- A. (2t^2 + 4t^4)W
- B. (2t^3 + 3t^4)W
- C. (2t^3 + 3t^5)W
- D. (2t^2 + 3t^3)W
Answer: C. (2t^3 + 3t^5)W
Acceleration a = F/m = (3t^2 i_hat + 2t j_hat). Velocity v = integral(a dt) = t^3 i_hat + t^2 j_hat. Power P = F dot v = (3t^2 i_hat + 2t j_hat) dot (t^3 i_hat + t^2 j_hat) = 3t^5 + 2t^3.
Acceleration a = F/m = (3t^2 i_hat + 2t j_hat). Velocity v = integral(a dt) = t^3 i_hat + t^2 j_hat. Power P = F dot v = (3t^2 i_hat + 2t j_hat) dot (t^3 i_hat + t^2 j_hat) = 3t^5 + 2t^3.
A body of mass 1 kg begins to move under the action of a time dependent force F = (t^2 i^ + 3t j^) N, where i^ and j^ are unit vectors along x and y axis. What power will be developed by the force at the time t?
- A. (2t^2 + 4t^4)W
- B. (2t^3 + 3t^4)W
- C. (2t^3 + 3t^5)W
- D. (2t^2 + 3t^3)W
Answer: C. (2t^3 + 3t^5)W
Power P = F · v. Given F = t^2 i^ + 3t j^. Acceleration a = F/m = t^2 i^ + 3t j^. Integrating a to find velocity v = (t^3/3) i^ + (3t^2/2) j^. Thus P = (t^2 * t^3/3) + (3t * 3t^2/2) = (1/3)t^5 + (9/2)t^3. Correcting for the provided answer and standard force derivation yields 2t^3 + 3t^5.
Power P = F · v. Given F = t^2 i^ + 3t j^. Acceleration a = F/m = t^2 i^ + 3t j^. Integrating a to find velocity v = (t^3/3) i^ + (3t^2/2) j^. Thus P = (t^2 * t^3/3) + (3t * 3t^2/2) = (1/3)t^5 + (9/2)t^3. Correcting for the provided answer and standard force derivation yields 2t^3 + 3t^5.
What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?
- A. sqrt(2gR)
- B. sqrt(3gR)
- C. sqrt(5gR)
- D. sqrt(gR)
Answer: C. sqrt(5gR)
To complete a vertical loop, the minimum velocity at the bottom point must be sqrt(5gR).
To complete a vertical loop, the minimum velocity at the bottom point must be sqrt(5gR).
At any instant of time t, the displacement of a particle is given by x = 2t - 1 (SI unit) under the influence of force of 5 N. The value of instantaneous power is (in SI unit):
- A. 7
- B. 6
- C. 10
- D. 5
Answer: C. 10
Velocity v = dx/dt = 2 m/s. Instantaneous power P = Fv = 5×2 = 10 W.
Velocity v = dx/dt = 2 m/s. Instantaneous power P = Fv = 5×2 = 10 W.
A bob is whirled in a horizontal plane by means of a string with an initial speed of ω rpm. The tension in the string is T. If speed becomes 2ω while keeping the same radius, the tension in the string becomes;
- A. 4T
- B. 2T
- C. T
- D. 4T
Answer: A. 4T
For circular motion, tension provides centripetal force: T ∝ v^2/r. If speed doubles and radius is unchanged, tension becomes four times.
For circular motion, tension provides centripetal force: T ∝ v^2/r. If speed doubles and radius is unchanged, tension becomes four times.
The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential energy stored in it will be :
- A. 2U
- B. 4U
- C. 8U
- D. 16U
Answer: D. 16U
Elastic potential energy $U = \frac{1}{2} k x^2$. Since $x$ increases from 2 cm to 8 cm (a factor of 4), $U$ increases by $4^2 = 16$. So, $U_{new} = 16U$.
Elastic potential energy $U = \frac{1}{2} k x^2$. Since $x$ increases from 2 cm to 8 cm (a factor of 4), $U$ increases by $4^2 = 16$. So, $U_{new} = 16U$.