Question

The initial speed of a 2.17-kg box traveling up a plane inclined 37° to the horizontal is 3.23 m/s. The coefficient of kinetic friction between the box and the plane is 0.30.

(a) How far along the incline does the box travel before coming to a stop?
m

(b) What is its speed when it has traveled half the distance found in Part (a)?
m/s

Answer 1

There are three forces acting on the block: mg is the gravitational force of the Earth, N is the normal force of the plane, F_k is the force of kinetic friction. The free-body diagram of the block is shown below:

The block is not allowed to move in direction normal to the plane, the net force on the block in this direction must be zero

The force of kinetic friction on the block is

The net force along the plane, decelerates the plane. By Newton's second law, the acceleration of the block is given by

we have taken the direction up the plane as positive and the direction down the plane to be negative

Let d be the distance traveled by the block before coming to a stop.

v_initial=3.23m/s, v_final=0, , d=?

Using

(b)

We want to find the speed when the distance traveled is

d₂=d/2=0.316m

Using