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

50.4 m

Step-by-step explanation:

We are given that

Coefficient of static friction =
`\mu_s=0.32`

Initial speed=u=64 km/h=
`64* (5)/(18)=17.78 m/s`

`1km/h=(5)/(18) m/s`

Final speed=v=0

`ma=-\mu mg`

`a=-\mu g=-0.32* 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=(-(17.78)^2)/(-2(3.136))`

`s=50.4 m`