Respuesta :
Explanation:
Mass of the clock, m = 95 kg
The horizontal force acting on the clock to set it in motion, [tex]F=650\ N[/tex]
A horizontal force to keep it moving is, F' = 560 N
Let [tex]\mu_s[/tex] is the coefficient of static friction. The horizontal force acting on the clock to set it in motion is given by :
[tex]F=\mu_sN[/tex]
[tex]F=\mu_s\times mg[/tex]
[tex]\mu_s=\dfrac{F}{mg}[/tex]
[tex]\mu_s=\dfrac{650}{95\times 9.8}[/tex]
[tex]\mu_s=0.69[/tex]
Let [tex]\mu_k[/tex] is the coefficient of static friction. The force due to motion of the clock is given by :
[tex]F'=\mu_kN[/tex]
[tex]F'=\mu_k\times mg[/tex]
[tex]\mu_k=\dfrac{F'}{mg}[/tex]
[tex]\mu_k=\dfrac{560}{95\times 9.8}[/tex]
[tex]\mu_k=0.60[/tex]
Hence, this is the required solution.
(A)The [tex]\mu_s[/tex] between the clock and the floor is 0.69
(B)The [tex]\mu_k[/tex] between the clock and the floor is 0.60
Explanation:
Given information:
The mass of the clock m [tex]=95[/tex] kg
The horizontal force [tex]F=650\;N[/tex]
The horizontal force to keep it moving [tex]F'=560\;N[/tex]
Suppose, [tex]\mu_s[/tex] is the coefficient of the static friction
Then the horizontal force will be:
[tex]F=\mu_s\times N\\F=\mu_s\times mg\\\mu_s=F/mg[/tex]
On putting the values in above equation:
[tex]\mu_s=0.69[/tex]
Let, [tex]\mu_k[/tex] is the coefficient of friction then
Force due to motion is given by:
[tex]F'=\mu_k\times mg\\\mu_k=F'/mg\\\mu_k=560/(95\times 9.8)\\\mu_k=0.60[/tex]
Hence,
The [tex]\mu_s[/tex] between the clock and the floor is 0.69
The [tex]\mu_k[/tex] between the clock and the floor is 0.60
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