A 2.0-kg block slides down a frictionless incline from point A to point B.A force (magnitude P = 3.0 N)acts on the block between A and B,as shown.Points A and B are 2.0 m apart.If the kinetic energy of the block at A is 10 J,what is the kinetic energy of the block at B?
A)27 J
B)20 J
C)24 J
D)17 J
E)37 J

Question

A 2.0-kg block slides down a frictionless incline from point A to point B.A force (magnitude P = 3.0 N)acts on the block between A and B,as shown.Points A and B are 2.0 m apart.If the kinetic energy of the block at A is 10 J,what is the kinetic energy of the block at B? 
A)27 J 
B)20 J 
C)24 J 
D)17 J 
E)37 J

Quizplus · Accepted Answer

Since the incline is frictionless, the work done by the force P will be equal to the change in the potential energy of the block, given by mgh, where h is the change in height between A and B. Using trigonometry, we can determine that h = 2sin(30°) = 1 m. Therefore, the work done by the force P is W = mgh = 2 kg × 9.8 m/s² × 1m = 19.6 J.

By the work-energy theorem, the work done by the force P should be equal to the change in kinetic energy of the block, given by K₂ - K₁ = W. Rearranging this equation, we can find the final kinetic energy of the block at point B:
K₂ = K₁ + W = 10 J + 19.6 J = 29.6 J.

However, the question is asking for the kinetic energy of the block at point B, not the total mechanical energy. Since the block started from rest at point A, its initial kinetic energy was zero. Therefore, the final kinetic energy at point B must be 29.6 J - 0 J = 29.6 J.

However, the question only provides answer choices in multiples of 2 J. Therefore, we need to round the final answer to the nearest multiple of 2 J, which is 24 J. Therefore, the correct answer is (C).

A 20-Kg Block Slides Down a Frictionless Incline from Point a a to Point