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asambeis [7]
3 years ago
13

A box of mass 20kg is pulled up an inclined plane by a force of 285N. Given that the value of the incline angle is 30 degrees an

d the coefficient of dynamic friction is 0.72, what is the speed with which the box is moving with, assuming it takes 4seconds to reach the top of the incline?
Physics
1 answer:
docker41 [41]3 years ago
6 0

Given :

Mass of box , m = 250 kg.

Force applied , F = 285 N.

The value of the incline angle is 30°.

the coefficient of dynamic friction is \mu=0.72 .

To Find :

The speed with which the box is moving with, assuming it takes 4 seconds to reach the top of the incline.

Solution :

Net force applied in box is :

F=285 - mgsin\ \theta - \mu mg cos \ \theta\\ \\F=285-mg( sin \ \theta - \mu cos\ \theta)\\\\F=285 - 20\times 10( \dfrac{1}{2}+0.72\times \dfrac{\sqrt{3}}{2})\\\\F=60.29\ N

Acceleration , a=\dfrac{F}{m}=\dfrac{60.29}{20}=3.01\ m/s^2.

By equation of motion :

v=u+at\\\\v=0+3.01\times 4\\\\v=12.04\ m/s

Therefore, the speed of box is 12.04 m/s.

Hence, this is the required solution.  

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Suppose a rocket ship in deep space moves with constant acceleration equal to 9.80 m/s2, which gives the illusion of normal grav
DochEvi [55]

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