Work-energy-power practice · Mixed difficulty

The correct answer is B. Potential energy is also referred to as mutual energy because it represents the energy of interaction between objects or particles.

Option A (30) is correct because the work done is proportional to the square of the distance stretched, so stretching an additional 's' requires 3 times the initial work.

The correct answer is D. A stretched bow possesses potential energy due to its deformation, which is stored and ready to be converted into kinetic energy.

None is correct because the potential energy is proportional to the square of the stretch, and stretching it 5 times more (from 2mm to 10mm) will result in 25 times more potential energy, which is 100J.

Option D is correct because in an inelastic collision, total energy and momentum are conserved, but kinetic energy is not conserved.

Option C (0.5) is correct because it is the ratio that satisfies the conditions of elastic collision and the given velocities.

Option A (32) is correct because the centripetal acceleration is calculated using $a = v^2/r$, where v is the speed and r is the radius.

Option D (98 J) is correct because the work done is equal to the change in potential energy, which is mgh, where h is the vertical height gained.

Option A (Work done by the boy against the gravitational force acting on the block) is correct because the gain in gravitational potential energy is equal to the work done against gravity.

Option A (Top of the circle) is correct because the tension is minimum at the top of the circle for a constant speed.

Option A (20) is correct because it is calculated by finding the force vector from the potential energy function and evaluating it at the given point.

Option B (-0.6 J) is correct because the change in mechanical energy is the difference between the initial mechanical energy and the final mechanical energy, which is negative due to energy loss from penetration into the sand.

Option D (Both (2) and (3)) is correct because rotating the frame of reference does not change the vector itself, but its components may change.

Option D is correct because work done by non-conservative forces depends on the path taken, not just the initial and final states.

Option A (If both Assertion and Reason are true and Reason is correct explanation of Assertion) is correct because the potential energy of a spring indeed varies parabolically with extension or compression, and this is because the potential energy is proportional to the square of the extension or compression.

Option A (1) is correct because the coefficient of restitution for a perfectly elastic collision is 1, indicating no energy loss.

Option A (Only momentum is conserved) is correct because, by definition, in an inelastic collision, kinetic energy is not conserved, but momentum is always conserved.

The correct answer is A (2 Mgr) because the difference in kinetic energy is equal to the difference in potential energy.

Option A (If both Assertion and Reason are true and Reason is correct explanation of Assertion) is correct because the Work-Energy Theorem states that the net work done on a particle equals its change in kinetic energy, and this applies directly to a single-particle system.

Option D (If the assertion is false and reason true) is correct because it is possible to accelerate while traveling at constant speed (e.g., circular motion), and it's impossible to round a curve with zero acceleration.

Option D (Statement I is false; Statement II is true) is correct because kinetic energy is minimum at the highest point in a vertical circle, and velocity is constant in magnitude for uniform circular motion in a horizontal plane.

Option C (${n^2}U$) is correct because the potential energy is proportional to the square of the extension, so increasing the extension by a factor of 'n' increases the potential energy by a factor of n squared.

Option A (Statement-I & II are true) is correct because engine B does the same work in less time, hence has more power, and power is indeed the dot product of force and velocity.

Option A is correct because in uniform circular motion, the force (tension in the string) is always perpendicular to the displacement, so no work is done.

Option B (31.4 m/s) is correct because average speed is total distance divided by total time, and for constant speed, it equals the speed itself.

Option D is correct because in a circular path with constant velocity, the net work done is zero as there is no change in kinetic energy.

Option A (1.4 km) is correct because it is calculated by finding the radius of the curve using the centripetal force condition and then determining the arc length for a 90-degree turn.

Option D (All of the above) is correct. Since KE is the same, and the retarding force is the same, the work done to stop them is the same, hence the distance is the same. Also, momentum depends on mass, and time to stop depends on mass.

The correct answer is B. The extra power required is equal to the rate of change of kinetic energy of the sand. P = d/dt (1/2 * m * v²) = v² * dm/dt = (6 m/s)² * 5 kg/s = 180 W.

Option C ($K/2$) is correct because at the highest point, the vertical velocity is zero, and only the horizontal component of the initial velocity remains, which is $v_0 \cos(45^\circ) = v_0/\sqrt{2}$. The kinetic energy at the highest point is therefore $K/2$.