A 2.0-N force acts horizontally on a 10-N block that is initially at rest on a horizontal surface. The coefficient of static friction between the block and the surface is 0.50. What is the magnitude of the frictional force that acts on the block?

Question

Makaila Freeman

Physics

1 June, 04:18

A 2.0-N force acts horizontally on a 10-N block that is initially at rest on a horizontal surface. The coefficient of static friction between the block and the surface is 0.50. What is the magnitude of the frictional force that acts on the block?

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Answers (1)

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Rosales · Answer 1

The force of friction is 2 N. The object isn't moving.

Explanation:

The frictional force on a object is given by it's normal force multiplied by the coefficient of static friction between the object and the surface. In this case since the surface is horizontal, the normal force is equal to the weight of the block with the oposite direction. So we have:

F_friction = 0.5*N = 0.5 * (10) = 5 N

Since the friction force is oposite to force that's acting on the block and it's greater than that force, the object won't move and the real force will be the same as the applied force. So we have:

F_friction = 2 N