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netineya [11]
3 years ago
15

A 250-kg crate is on a rough ramp, inclined at 30° above the horizontal. The coefficient of kinetic friction between the crate a

nd 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?

Physics
2 answers:
astraxan [27]3 years ago
6 0

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

scoundrel [369]3 years ago
6 0

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

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