Question

A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back of a truck. Thecoefficient of static friction between box andtruck bed is 0.202.The acceleration of gravity is 9.8 m/s2 .When the truck accelerates forward, whatforce accelerates the box?006 (part 2 of 2)
What maximum acceleration can the truck
have before the box slides?
Answer in units of m/s2.

Answer 1

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity=$9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates=$\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates=$62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide=$1.9796 m/s^2$