A 100 N box is initially at rest on a rough horizontal surface. The coefficient

Question

A 100 N box is initially at rest on a rough horizontal surface. The coefficient of static friction between the box and the surface. The coefficient of static friction between the box and the surface is 0.6 and the coefficient of kinetic friction is 0.4 A constant 35 N force is applied to the box horizontally as shown. Identify from choices (a) - (e) how each change described below will affect the frictional force on the box by the surface I second after the horizontal force is first applied Compared to the case above, this change will: (a) increase the friction force exerted on the box by the surface. (b) decrease the friction force exerted on the box by the surface but not on zero. (c) decrease the friction force exerted on the box by the surface to zero. (d) have no effect on the friction force exerted on the box by the surface. (e) have an indeterminate effect on the friction force exerted on the box by the surface. All of these modification are change to the initial situation shown in the diagram. 1) The weight of the box is change to 50 N. 2) The weight of the box is change to 200 N. 3) The applied force is increased to 50 N. 4) The applied force is increased to 80 N. 5) The coefficient of static friction is increased to 0.7. 6) The coefficient of kinetic friction is increased to 0.5. 7) The coefficient of kinetic friction is increased to 0.5 and the coefficient of friction is increased to 0.7. 8) The weight of the box is change to 200 N and the coefficient of static friction is increased to 0.7. 9) The weight of the box is changed to 200 N and the coefficient of kinetic friction is increased to 0.5. 10) The weight of the box is change to 200 N and the applied force is increased to 50 N.

Explanation / Answer

Frictional force on the block is f = µN

where µ = coefficient of friction between the block and the surface and

N = normal force exerted on the block by the surface

Since there is no vertical motion, the vertical accelration is zero and so is the sum of vertical forces which makes

N = mg (where m = mass of the block and thus, mg is the weight of the block)

Initially the weight of the block is W = 100N

So maximum static friction which the surface can apply before the block moves is:

fs = µsN = 0.6(100N) = 60N

So initially, the block won't move as the applied force is F = 35N

So initially, the fritional force on the block is f = 35N (block has zero accln. so, F = f)

Now let us consider the modifications:

1. The weight of the box is canged to 50N.

In this case maximum static friction which the surface can apply before the block moves is:

fs = µsN = 0.6(50N) = 30N

but the applied force is 35N, hence the block starts moving if the weight is changed to 50N. But now the frictional force on the block will be kinetic friction for which µk = 0.4

So force of kinetic friction is fk = µkN = 0.4(50N) = 20N

So, the frictional force will be decreased if the weight is changed to 50N.

  

2. The weight of the box is canged to 200N.

In this case the maximum static friction which the surface can apply before the block moves is:

fs = µsN = 0.6(200N) = 120N

but the applied force is F = 35N. So the block won't move. So, the frictional force will be f = 35N (block has zero accln. so, F = f).

So there will be no change in the frictional force, if the weight is changed to 200N

3.the weight of the block is W = 100N

So maximum static friction which the surface can apply before the block moves is:

fs = µsN = 0.6(100N) = 60N

So the block won't move as the applied force is F = 50N

So , the fritional force on the block is f = 50N (block has zero accln. so, F = f)

So, the frictional force will be increased if the applied force is changed to 50N.

4. the weight of the block is W = 100N

So maximum static friction which the surface can apply before the block moves is:

fs = µsN = 0.6(100N) = 60N

So the block will move as the applied force is 80N. When the block moves, the frictional force on the block will be kinetic friction for which µk = 0.4

So force of kinetic friction is fk = µkN = 0.4(100N) = 40N

So the frictional force will increase if applied force is 80N.

This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification or correction I will be happy to oblige....

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