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3 years ago

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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?

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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