Two blocks are sliding together along a frictionless surface. Block A, on the le

Question

Two blocks are sliding together along a frictionless surface. Block A, on the left, has a mass of 1.0 kg and block B, on the right, has a mass of 7.0 kg. Another object is exerting a force on B. The horizontal velocity as a function of time of both A and B is shown in the graph below (right is the positive direction). What is the horizontal force exerted by B on A? What is the force exerted on B by A? What is the force exerted by the other object on B? (Be sure to indicate direction with the sign.)

Force exerted by B on A =
Force exerted on B by A =
Force exerted on B by other object =

Explanation / Answer

The graph shows initial velocity=+6 m/s that decreases linearly to +2 m/s after 5 s.

The system (w/ block A and block B) has a constant acceleration:

(2 m/s 6 m/s) / 5 s = -0.8 m/s²

Using Newton's Second Law:

Block A has a force of F=m*a=4 kg * -0.8 m/s² =3.2 N

Because the surface is frictionless, the only force on block A is the one exerted by block B, which is -3.2 N.

By Newton's Third Law, the force exerted by block A on block B is equal and opposite to that exerted on block A by block B, so B exerts a force of +3.2 N on A.

By Newton's Second Law, the total force exerted on block B is F = mA= 5 kg * -0.8 m/s²= -4.0 N. This is the sum of the force of block A and the external object X on block B. Therefore
-4.0 N = 3.2 N + X
the external object therefore exerts a force of
X = -4.0 + -3.2 N = -7.2 N on block B.

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