A 20.0 kg crate sits at rest at the bottom of a 18.0 m long ramp that is incline

Question

A 20.0 kg crate sits at rest at the bottom of a 18.0 m long ramp that is inclined at 32 degrees above the horizontal. A constant horizontal force of 290 N is applied to the crate to push it up the ramp. While the crate is moving, the ramp exerts a constant frictional force on it that has magnitude 65.0 N.

a)

What is the total work done on the crate during its motion from the bottom to the top of the ramp?

Express your answer to three significant figures and include the appropriate units.

b)

How much time does it take the crate to travel to the top of the ramp?

Express your answer to three significant figures and include the appropriate units.

All Yahoo answers are wrong, please help.

Explanation / Answer

Here ,

m   = 20 Kg

d = 18 m

theta = 32 degree,

F = 290 N

f = 65 N

Total work done = WF + Wfricition + Wgravity

Total work done = F*d*cos(theta) - f*d - mgd*sin(theta)

Total work done = 290*18*cos(32) - 65 * 18 - 20 *9.8*18*sin(32)

Total work done = 1387.3 J

the total work done on the crate is 1387.3 J

b)

Now , Here , along the surface of incline

Fnet = F*cos(theta) - f - mg*sin(theta)

Fnet = 290 - 65 - 20*9.8 *sin(32)

Fnet = 121.2 N

Using second law of motion ,

Fnet = ma

a = 121.2/20

a = 6.06 m/s^2

Now , using second equation of motion

d = ut + 0.5 at^2

18 = 0.5 * 6.06 * t^2

t = 2.44 s

the time taken to reach the top of incline is 2.44 s

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