# A box rests on a truck moving with a velocity of 2 m/s. The coefficient of friction between box and truck is 0.1 …..

Q: A box rests on a truck moving with a velocity of 2 m/s. The coefficient of friction between box and truck is 0.1. Driver apply the brake and truck starts to decelerate uniformly and stops in 0.5 s. The total distance travelled by the box with respect to truck is

(a) 0.75 m

(b) 0.375 m

(c) 1.5 m

(d) 2 m

Ans: (c)

Sol: Distance travelled by truck is

$\displaystyle S_1 = \frac{2+0}{2}\times 0.5$

S1 = 0.5 m

Acceleration of the box is

$\displaystyle a = \frac{\mu m g}{m} = \mu g$

a = 0.1 × 10 = 1 m/s2

v2 = u2 + 2 a S2

0 = 22 – 2 × 1 × S2

S2 = 2 m

Hence , Distance travelled by box with respect to truck is

= S2 – S1

= 2 – 0.5 = 1.5 m