1 A box is at rest on a horizontal floor A force FF acts hor

1) A box is at rest on a horizontal floor. A force FF acts horizontally on the side of the box, but static friction between the floor and the box keeps it from moving. Which of the following choices would increase the magnitude of the static frictional force acting between the box and the floor? Assume that the block does not begin to slide.

(a) Increasing the applied force FF.

(b) Increasing the static coefficient of friction between the box and the floor.

(c) Increasing the weight of the box.

(d) Either or both of choices (b) or choice (c).

(e) None of the above.

2) Suppose you are pushing a box across a horizontal floor at a constant velocity V1V1. There is friction between the box and the floor, and the magnitude of the horizontal force that you are applying to the side of the box to keep it moving is F1F1. Suppose your friend is pushing an identical box across the same floor at a higher constant velocity V2V2, and that the magnitude of the horizontal force that your friend is applying to the side of her box to keep it moving is F2F2. Compare F1F1 and F2F2.

(a) F1<F2F1F2

(b) F1=F2F1F2

(c) F1>F2

3) A box is pushed across a horizontal floor in the x-direction with a some initial velocity. At t = 0 force is removed, after which the box slows down and stops because of kinetic friction between the floor and the box. The coefficient of kinetic friction between the box and the floor is KK. Which of the following graphs best describes the acceleration of the box after the force is removed ?

A

B

C

D

E

Solution

( 1 )

Expression for static frictional force:

F_s = _s F_n

Here, F_s is static friction,s is the coefficient of static friction and F_n is the normal force.

From the above equation, it is cleared that increasing the static coefficient of friction between the box and the floor. or increasing the weight of the box would results in increaing static frictional force.

Thus, the correct option is ( d )

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( 2)

Acceleration of object on the horizontal surface can be written as ( by using Newton\'s second law )

a = ( F - f ) / m

a = ( F - mg ) / m

a = F /m - g

In order to move with constant velocity , acceleration should be constant. For maintain high contnat acceleration, the force should be higher and constant.

Hence, for the given situation, F1 < F2

Thus, the correct option is ( a )

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( c )

In this case, the velocity of box gradully decreaes at constant rate. Therefore, the deceleration of box is constnat. The graph ( c) represents the given situation.