Question

A plank with a box on it at one end is gradually raised about the other end. As the angle of

inclination with the horizontal reaches 30° , the box starts to slip and slides 4.0 m down the

plank in 4.0 s. The coefficients of static and kinetic friction between the box and the plank will
be, respectively ?

#IIT-JAM​

Answer 1

Answer:

ANSWER

Static coefficient of friction is μ s= tan 30°= 0.577≈0.6

For kinetic friction, ma=mgsin30−f=mgsin30−μkmgcos 30

a= gain 30-μ k g cos 30....(1)

and also using S=ut+1/2at.sq

⇒4=0+(1/2)a(4) ^2

or a=0.5m/s ^2

Now from (1) we get, 0.5=10(1/2)−μ k(10)(√3/2)

or μ k= 4.5/5√3

=0.5

Answer 2

0.5 and 0.6

Explanation:

Static coefficient of friction is μ>s

tan30° =0.577≈0.6

For kinetic friction, ma=mg sin30−f=mg sin30−μ>k- mg cos30

a=gsin30−μ>k g cos30....(1)

Or using s=ut+1/2at²

⇒4=0+(1/2)a(4)

or a=0.5m/s²

Now from equation (1) we get, 0.5=10(1/2)−μ>k

μ>k(10)(√3/2)

or μ>k=4.5/5√3=0.5

This is your answer mate.!

Hope this helps you!