A block of mass 2.5 mg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is found that it takes 5 seconds to slide through the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion.Taking block is gently pushed to start the motion, calculate the coefficient of static and kinetic friction between the block and the surface. Taking `g=10 m//s^2`

Options

A. `mu_s = 0.52` , `mu_k = 0.60`
B. `mu_s = 0.60` , `mu_k = 0.52`
C. `mu_s = 0.60` , `mu_k = 0.72`
D. `mu_s = 0.72` , `mu_k = 0.60`

Answer

B

Explanation

The forces acting on the block are shwon in ure. Here M=25 kg and F=15N`.

When F=15N is applied to the block remains in limiting equilibrium. The force of friction is thus `f=mu_sN`. Applying Newton's first law.
`f=mu_sN and N=mg`
so that `Fmu_s Mg`
or, `mu_s=F/mg= (15N)/((25 kg)(10m//s^2))=0.60`
when the block is gently pushed to start the motion, kinetic friction acts between the block and the surface. Since the block takes 5 second to slide through the first 10 m, the acceleration a is given by
`10 m = 1/2 a(5s)^2`
or, `a=20/25 m//s^2 =0.8m//s^2`
The frictional force is `
f=mu_kN=mu_kMg`
Applying Newton's second law
`F-mu_k Mg=Ma`
or `mu_k=(F-Ma)/Mg`
` (15N-(2.5 kg)(0.8 m//s^2))/((2.5 kg)(10 m//s^2))=0.52`