9702_s22_qp_12 Question 8

A child of mass 20 kg stands on the rough surface of a sledge of mass 40 kg. The sledge can slide on a horizontal frictionless surface. 

One end of a rope is attached to the sledge. The rope passes around a fixed frictionless pole, and the other end of the rope is held by the child, as shown. 

The rope is horizontal. The child pulls on the rope with a horizontal force of 12 N. This causes the child and the sledge to move with equal acceleration towards the pole. 

What is the frictional force between the child and the sledge? 

A 4.0 N 

B 6.0 N 

C 8.0 N 

D 12 N  

ANSWER: A

The Physics Behind

• From the statement/given information "This causes the child and the sledge to move with equal acceleration towards the pole.", the following should be satisfied:
• For the sledge

a = F_s / m_T

where  F_s = 12 N + 12 N and m_T = 60 kg

F_s is the resultant force on sledge

m_T is the total mass

• For the child

a = F_c  / m_c

where F_c = 12 - f and m_c = 20 kg

F_c  is the resultant force on the child

m_c is the mass of the child

f is the frictional force

• The accelerations are equal so

0.40 N kg-1 = (12 N - f) / 20 kg

8.0 N = 12 N - f

f = 8.0 N - 12 N

f = 4.0 N