A 4.10kgblock is pushed along a floor by a constant applied force that is horizontal and has a magnitude of 40.0N. Figure 6-30 gives the block’s speed vversus time tas the block moves along an xaxis on the floor. The scale of the figure’s vertical axis is set by vs=5.0m/s. What is the coefficient of kinetic friction between the block and the floor

Short Answer

Expert verified

The coefficient of kinetic friction between the block and the floor is 0.53.

Step by step solution

01

Given

Mass of the block,m=4.10kg

Force,F=40.0N

02

Determining 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. To initiate, the acceleration of the block can be found from the graph and then the free body diagram for the block can be drawn. By using Newton's 2nd law of motion, coefficient of friction between the table and box can be found.

Formula:

Fnet=ma

where, F is the net force, mis mass and a is an acceleration.

03

Determining the free body diagram

Free Body Diagram of the system:

04

Determining the coefficient of kinetic friction between the block and the floor

Acceleration of the block from graph,

a=vt=(5.0)-(0.5)(1.0)-(0)=4.5m/s2

By using Newton’s 2nd law of motion along the horizontal direction,

F-fk=mafk=F-mafk=40.0-(4.10)(4.5)=21.55N(i)

By using Newton’s 2nd law of motion along the vertical direction,

N+mg=0N=mg

Thus, the frictional force,

fk=μsN=μk(mg) (ii)

From Eq. (1) and (2),

21.55=μk(mg)

μk=21.55mg=21.55(4.10)(9.81)=0.53

Hence, the coefficient of kinetic friction between the block and the floor is 0.53.

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