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A child pushes her toy across a level floor at a steady velocity of 0.50 m/s using an applied force of 2.0 N. If the weight of the toy is 8.0 N, what is the coefficient of sliding friction between the toy and the floor?

A0.25
B1.0
C4.0
D0.40

Verified Answer :

Lampa20

since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.

Rate of speed that toy is moving is irelevant.

childs force is:
Fc = 2N
Fc = Ff  (Ff -friction force)

Ff = a*Q

where Q is weight of the toy and a is friction

if we express a we get
a = F/Q = 2/8 = 0.25

skyluke89

Answer:

a. 0.25

Explanation:

The toy is moving across the floor at constant velocity - this means that its acceleration is zero.

According to Newton's second law, this means that the net force acting on the toy is zero:

\(F_{net}=ma=0\) (1)

because a=0.

The net force, in this case, consists of two forces acting in opposite directions:

- The applied force, F = 2.0 N

- The frictional force, \(F_f = -\mu (mg)\)

where \(\mu\) is the coefficient of sliding friction and (mg)=8.0 N is the weight of the toy.

Therefore, the equation of the forces (1) becomes:

\(F+F_f = 0\\F-\mu (mg)=0\\\mu = \frac{F}{mg}=\frac{2.0 N}{8.0 N}=0.25\)

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