The surface of the inclined plane shown is frictionless. If F = 30 N, what is the magnitude of the force exerted on the 3.0-kg block by the 2.0-kg block?
A) 18 N
B) 27 N
C) 24 N
D) 21 N
E) 15 N

Question

The surface of the inclined plane shown is frictionless. If F = 30 N, what is the magnitude of the force exerted on the 3.0-kg block by the 2.0-kg block? 
A) 18 N 
B) 27 N 
C) 24 N 
D) 21 N 
E) 15 N

Quizplus · Accepted Answer

Since the surface is frictionless, the only forces acting on the blocks are gravity and tension. The 3.0-kg block has a weight of 29.4 N (3.0 kg × 9.8 m/s²). The 2.0-kg block has a weight of 19.6 N (2.0 kg × 9.8 m/s²).

The tension force is the same for both blocks since they are connected by the same rope. Therefore, we can write an equation for each block based on their acceleration:

For the 3.0-kg block: T - 29.4 N = (3.0 kg) a

For the 2.0-kg block: T - 19.6 N = (2.0 kg) a

Since we want to find the force exerted on the 3.0-kg block by the 2.0-kg block, we can subtract the second equation from the first to eliminate T and solve for a:

9.8 N = (3.0 kg + 2.0 kg) a

a = 1.96 m/s²

Now that we know the acceleration of the blocks, we can use either equation to solve for T:

T = 29.4 N + (3.0 kg)(1.96 m/s²) = 35.3 N

T = 19.6 N + (2.0 kg)(1.96 m/s²) = 23.5 N

Since the tension force is the same for both blocks, we can use either value to find the force exerted on the 3.0-kg block by the 2.0-kg block:

F = T - 19.6 N = 35.3 N - 19.6 N = 15.7 N

Therefore, the answer is A) 18 N (which is closest to 15.7 N).

The Surface of the Inclined Plane Shown Is Frictionless