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vodka [1.7K]
3 years ago
13

A 95 kg clock initially at rest on a horizontal floor requires a 650 n horizontal force to set it in motion. after the clock is

in motion, a horizontal force of 560 n keeps it moving with a constant velocity.a) what is the μs between the clock and the floor?.70b) what is the μk between the clock and the floor?
Physics
2 answers:
iogann1982 [59]3 years ago
8 0

Explanation:

Mass of the clock, m = 95 kg

The horizontal force acting on the clock to set it in motion, F=650\ N

A horizontal force to keep it moving is, F' = 560 N

Let \mu_s is the coefficient of static friction. The horizontal force acting on the clock to set it in motion is given by :

F=\mu_sN

F=\mu_s\times mg

\mu_s=\dfrac{F}{mg}

\mu_s=\dfrac{650}{95\times 9.8}

\mu_s=0.69

Let \mu_k is the coefficient of static friction. The force due to motion of the clock is given by :

F'=\mu_kN

F'=\mu_k\times mg

\mu_k=\dfrac{F'}{mg}

\mu_k=\dfrac{560}{95\times 9.8}

\mu_k=0.60

Hence, this is the required solution.

Tamiku [17]3 years ago
3 0
The μs between the clock and floor is 650(M*g) and the μk between the clock and the floor is 560(M*g)
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