8 A child of mass 20 kg stands on the rough surface of a sledge of mass 40 kg. The sledge canslide 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.horizontal rope frictionlesspolefrictionless surfaceThe rope is horizontal. The child pulls on the rope with a horizontal force of 12 N. This causes thechild 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
Question
8 A child of mass 20 kg stands on the rough surface of a sledge of mass 40 kg. The sledge canslide 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.horizontal rope frictionlesspolefrictionless surfaceThe rope is horizontal. The child pulls on the rope with a horizontal force of 12 N. This causes thechild 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
Solution
The child and the sledge move together with equal acceleration, which means the force applied by the child is distributed over the total mass, which is 20 kg (child) + 40 kg (sledge) = 60 kg.
The force applied (F) is equal to the mass (m) times the acceleration (a), or F = ma.
We can rearrange this formula to find the acceleration: a = F/m.
Substituting the given values, we find the acceleration: a = 12 N / 60 kg = 0.2 m/s².
The frictional force between the child and the sledge is the force that the child applies to the sledge. This force is equal to the mass of the child times the acceleration: F = ma.
Substituting the given values, we find the frictional force: F = 20 kg * 0.2 m/s² = 4 N.
So, the frictional force between the child and the sledge is 4.0 N.
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