Question

A student wants to determine the coefficients of static friction and kinetic friction between a box and a plank. She places the box on the plank and gradually raises one end of the plank. When the angle of inclination with the horizontal reaches 30°, the box starts to slip, and it then slides 2.5 m down the plank in 4.0 s at constant acceleration. What are (a) the coefficient of static friction and (b) the coefficient of kinetic friction between the box and the plank?

Answer 1

Answer:

Explanation:

Given

at $\theta =30^{\circ}$

Box slides down $2.5 m\ in\ t=4 s$

as box just starts to slide at $\theta =30^{\circ}$

friction force will just balance the sin component of weight

thus

$mg\sin \theta =f_r$

and $f_r=\mu N$

where $\mu$is coefficient of static friction

$N=mg\cos \theta$

thus $\mu =\tan \theta$

$\mu =0.577$

(b)block travels 2.5 m in 4 s

using $s=ut+\frac{at^2}{2}$

$2=0+\frac{a\times 16}{2}$

$a=0.25 m/s^2$

i.e. $a=\mu _kg$

$\mu_k=0.025$