Question

iii) A body of mass 37 kg rests on a rough
horizontal surface. The minimum
horizontal force required to just start the
motion is 68.5 N. In order to keep the
body moving with constant velocity, a
force of 43 N is needed. What is the value
of a) coefficient of static friction? and b)
coefficient of kinetic friction?
​

Answer 1

Answer:

The coefficient of static friction is 0.19

The coefficient of kinetic friction is 0.12

Explanation:

Mass of the body M = 37 kg

Minimum force required to start the motion F = 68.5 N

Before the motion has started, the coefficient of friction acting will be the static coefficient of friction

Thus,

F=\mu_s\times Mg

\implies \mu_s=\frac{F}{Mg}

\implies \mu_s=\frac{68.5}{37\times 9.8}

\implies \mu_s=0.19

When the body is in motion, the friction force acting will be kinetic friction

Given that the force acting to keep the motion with constant velocity F' = 43 N

Therefore, the coefficient of kinetic friction will be given by

F'=\mu_k\times Mg

\implies \mu_k=\frac{F}{Mg}

\implies \mu_k=\frac{43}{37\times 9.8}

\implies \mu_k=0.12

Hope this helps.

Answer 2

Solution :

As per the given data ,

• Mass of the body = 37 kg
• Minimum force required to start the motion = 68.5 N
• Force required to keep the body in motion = 43 N

(1)

We know that ,

→ Fs = μs N

→ Fs = μs x mg

here ,

• Fs = Minimum force applied on the body
• μs = static friction
• m = mass of the body
• g = acceleration due to gravity

→ μs = Fs / m x g

→ μs = 68.5 / 37 x 10

→ μs = 0.18

(2)

Similarly ,

→ Fk = μk x N

→ Fk = μs x mg

here ,

• Fs = Force applied to keep the body in motion
• μs = static friction
• m = mass of the body
• g = acceleration due to gravity

→ μk = Fk / m x g

→ μk = 43 / 37 x 10

→ μk = 0.12

Answer :

• The coefficient of static friction is 0.18
• The coefficient of kinetic friction is 0.12