A box is being pulled to the right over a rough surface. t > fk , so the box is speeding up. Suddenly the rope breaks. What happens? The box:. a. keeps its speed for a short while, then slows and stops. slows steadily until it stops.
b. stops immediately.
c. continues speeding up for a short while, then slows and stops.
d. continues with the speed it had when the rope broke.

a. keeps its speed for a short while, then slows and stops. slows steadily until it stops.

Explanation:

Since the tension in the rope, t is greater than the kinetic friction fk, the box is moving forward because there is a net force on it. That is, t - fk = f = ma.

Since there is a net force, there is an acceleration and thus an increasing velocity.

When the rope breaks, the tension, t = 0. So, t - fk = 0 - fk = -fk = ma'.

Now, the net force acting on the box is friction in the opposite direction. This force tends to slow the box down from its initial velocity at acceleration, 'a' until its velocity is zero, where it stops. Since the frictional force is constant, the acceleration, a' on the box is thus constant and the box undergoes uniform deceleration until its velocity is zero.

So, the box keeps its speed for a short while, then slows and stops. slows steadily until it stops.

So, the answer is a.

the area between the graph and the t axis