A block of mass 2.5 mg is kept on a rough horizontal surface. It is found tht the blocl 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 te block is gently pushed to start the motin.Taking block is gently pushed to start the motion. Taking `g=10 m/s^2`, calculate the coefficientof staticand kietic friction between the block and the surface.

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 teh motion, kinetic friction acts between the block and the surfce. Since the block takes 5 second to slide through teh 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 frictioN/Al 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))=050`