Question

A 5.340 kg5.340 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is ?s=0.455μs=0.455 and the coefficient of kinetic friction is ?k=0.155.μk=0.155. At time ?=0,t=0, a force ?=14.7 NF=14.7 N is applied horizontally to the block. State the force of friction applied to the block by the table at times ?=0t=0 and ?>0

Consider the same situation, but this time the external force ?F is 29.6 N.29.6 N. Again, state the force of friction acting on the block at times ?=0t=0 and ?>0.

Answer 1

Given:

Mass of the wooden block,

Coefficient of static friction,

Coefficient of kinetic friction,

Frictional force = Coefficient of friction*Normal force

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Normal force,

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Static frictional force,

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Kinetic frictional force,

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Given that the wooden block is resting on the steel desk.

In both cases, a horizontal force is applied to the wooden block. If the block is not moving, then the frictional force acting is the static frictional force. If the applied force is large enough to overcome the static frictional force, then the block will start moving and frictional force present will be the kinetic frictional force.

Consider the 1st case

Applied force,

In this case, the applied force is less than the static frictional force. Therefore the block will not move.

ANSWER: static frictional force.

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Consider the 2nd case

Applied force,

In this case, the applied force is larger than the static frictional force. The block will start moving and frictional force acting will be kinetic.

ANSWER: kinetic frictional force.

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