Question

A 1.5 kg box is initially at rest on a horizontal surface when at t =0 a horizontal force F = (1.8t) iN (With t in seconds), is applied to the box. The seconds), is applied to the box. The acceleration of the box as a function of time t is given by.
(g = $10m/s^2$)
$\overline{a}=0for0\le t\le 2.85$
$\overline{a}=(1.2t-2.4)\overline{j}m/s^{2}fort>2.85$
The coefficient of kinetic friction between the box and the surface is

• A. 0.12
• B. 0.24
• C. 0.36
• D. 0.48

Answer 1

Answer: (B)

0.24

m$=1.5kg$
coefficient of friction$=\mu$
the frictional force $f_k=\mu mg=15\mu N$
net force on the box $F=1.8t-15\mu$
acceleration of the box $a=\frac{f}{m}=\frac{1.8t-15\mu }{1.5}=1.2-10\mu$
$10\mu =2.4$
$\mu =0.24$