Question

The coefficient of static friction between the flat bed of the truck and the crate it carries is 0.32. Determine the minimum stopping distance s which the truck can have from a speed of 64 km/h with constant deceleration if the crate is not to slip forward.

Answer 1

Answer:

50.4 m

Explanation:

We are given that

Coefficient of static friction =$\mu_s=0.32$

Initial speed=u=64 km/h=$64\times \frac{5}{18}=17.78 m/s$

$1km/h=\frac{5}{18} m/s$

Final speed=v=0

$ma=-\mu mg$

$a=-\mu g=-0.32\times 9.8=-3.136 m/s^2$

Using $g=9.8 m/s^2$

We have to find the distance s.

$v^2-u^2=2as$

Using the formula

$0-(17.78)^2=-2(3.136)s$

$-(17.78)^2=-2(3.136)s$

$s=\frac{-(17.78)^2}{-2(3.136)}$

$s=50.4 m$