Question

A child of mass m slides down a slide inclined at 31.5° in time t1. The coefficient of kinetic friction between her and the slide is μk. She finds that if she sits on a small sled (also of mass m) with frictionless runners, she slides down the same slide in time 1/2t1. Find μk.

Answer 1

if we make the Free Body Diagram of the child along the slide, the forces are as follows:

mg sin(31.5) along the slide downwards

(1/4)k mg cos(31.5) along the slide upwards.

so net acceleration in the downward direction along the slide =

= g [sin(31.5) - (1/4)k cos(31.5) ] ...........(1)

Since by second equation of uniform acceleration motion

S = ut + (1/2) a*t^2

since u = 0

so, S = 1/2 a t^2

Since the same distance is travelled in time t and t/2 in the two cases respectively

so equating the displacements, we get:

1/2 g [sin(31.5) - (1/4)k cos(31.5) ] t1^2 = (1/2) g sin(31.5) *(t1/2)^2

So, sin(31.5) - (1/4)k cos(31.5) = sin(31.5)/4

So, (1/4)k = (3/4) tan(31.5) = 0.4595

k = 1.838