Question

A 250-kg crate is on a rough ramp, inclined at 30° above the horizontal. The coefficient of kinetic friction between the crate and ramp is 0.22. A horizontal force of 5000 N is applied to the crate, pushing it up the ramp. What is the acceleration of the crate?

Answer 1

The force that is pushing the crate up the ramp is competing with at least another two forces, those being G force= defines the attraction to the earth and Ff, the friction between the crate and the rough material of the ramp

Any Force can be defined by the weigh of a particular body and the acceleration

the kinetic friction indicative would be defined by the report between the active force and the friction force, considering G

So 5000-22%×5000=250× a

a=3900/250 m/s

Answer 2

Answer:

$8.35 m/s^2$

Explanation:

We are given that

Mass of crate=250 kg

$\theta=30^{\circ}$

Coefficient of kinetic friction between the crate and ramp=0.22

Horizontal force applied on the crate=5000 N

We have to find the acceleration of the crate.

The normal force acting on the crate

$=N=Fsin\theta+mgcos\theta=5000sin 30^{\circ}+250\times 9.8\times cos30^{\circ}=4623.93 N$

The friction force acting against the motion of crate=$0.22\times 4623.93=1017.26 N$

According to newton's second law, the net force accelerating the crate

$ma=Fcos\theta-(F_f+mgsin\theta)$

$a=\frac{5000cos 30^{\circ}-(1017.26+250\times 9.8 sin30^{\circ}}{250}$

$a=8.35 m/s^2$

Hence, the acceleration of the crate=$8.35 m/s^2$