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Chapter 6: Q6P (page 140)

A baseball player with massm=79kg, sliding into second base, is retarded by a frictional force of magnitude470N.What is the coefficient of kinetic frictionμkbetween the player and the ground?

Short Answer

Expert verified

The coefficient of kinetic friction between the player and the ground is 0.61.

Step by step solution

01

Write the given information:

Mass,m=79kg

Frictional force,fs=470N

02

Determine the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it.

Consider the formula for the second law of motion:

Fnet=∑ma

Here, F is the net force, mis mass and ais an acceleration

03

Determine thefree body diagram

Free body diagram of basketball player:

04

Determine the coefficient of kinetic friction

By using Newton’s 2nd law along y direction,

∑Fy=may

Since player is not moving along y.ay=0

N-mg=0N=mg

Relation between frictional force and Normal force is as follows:

fs=μkN

Rewrite the formulas, substitute the values and solve as:

μk=fsN=fsmg=470799.81=0.61

Hence, the coefficient of kinetic friction between the player and the ground is 0.61.

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Most popular questions from this chapter

A filing cabinet weighing 556 Nrests on the floor. The coefficient of static friction between it and the floor is 0.68, and the coefficient of kinetic friction is 0.56. In four different attempts to move it, it is pushed with horizontal forces of magnitudes (a) 222 N, (b) 334 N, (c) 445 N, and (d) 556 N. For each attempt, calculate the magnitude of the frictional force on it from the floor. (The cabinet is initially at rest.) (e) In which of the attempts does the cabinet move?

A certain string can withstand a maximum tension of 40 Nwithout breaking. A child ties a 0.37 kgstone to one end and, holding the other end, whirls the stone in a vertical circle of radius 0.91 M, slowly increasing the speed until the string breaks. (a) Where is the stone on its path when the string breaks? (b) What is the speed of the stone as the string breaks?

The two blocks(m=16kgandM=88kg)in Fig. 6-38 are not attached to each other. The coefficient of static friction between the blocks is μs=0.38,but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force Frequired to keep the smaller block from slipping down the larger block?

A person riding a Ferris wheel moves through positions at (1) the top, (2) the bottom, and (3) mid height. If the wheel rotates at a constant rate, rank these three positions according to (a) the magnitude of the person’s centripetal acceleration, (b) the magnitude of the net centripetal force on the person, and (c) the magnitude of the normal force on the person, greatest first.

A bicyclist travels in a circle of radius 25.0 mat a constant speed of 9.00 m/s. The bicycle–rider mass is 85.0 kg. Calculate the magnitudes of

(a) the force of friction on the bicycle from the road and

(b) the netforce on the bicycle from the road.
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