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netineya [11]
2 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]2 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]2 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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1. A diffraction grating with 5.000 x 103 lines/cm is used to examine the sodium
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7 1
2 years ago
Read 2 more answers
A golfer hits a golf ball at an angle of 25.0° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is
kvasek [131]

<u>Answer:</u>

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<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

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     We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

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Now we have H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m

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7 0
2 years ago
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