Question

The frictional force acting on a small block of mass 0.15 kg, while it is moving on a horizontal surface,
has magnitude 0.12 N. The block is set in motion from a point X on the surface, with speed 3 m s−1
. It
hits a vertical wall at a point Y on the surface 2 s later. The block rebounds from the wall and moves
directly towards X before coming to rest at the point Z (see diagram). At the instant that the block
hits the wall it loses 0.072 J of its kinetic energy. The velocity of the block, in the direction from X to
Y, is vm s−1
at time ts after it leaves X.

Answer 1

(i) Find the values of v when the block arrives at Y and when it leaves Y, and find also the value of
t when the block comes to rest at Z. Sketch the velocity-time graph. [9]
(ii) The displacement of the block from X, in the direction from X to Y, is s m at time ts. Sketch the
displacement-time graph. Show on your graph the values of s and t when the block is at Y and
when it comes to rest at Z

Answer 2

Apply Newtons second Law Sum of all forces in the direction of motion- sum of forces opposite to motion=ma
-0.12=0.15a
a=-0.12/0.15

Answer 3

So there is a negative acceleration, due to the resistive frictional force of 0.12N

Answer 4

@aaronq

Answer 5

use v=u+at
v=3+(-0.8(2))=1.4 m/s to approach the wall

Answer 6

@Kainui