- #1

DoJang

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**:( Please help! Work,power and energy question**

## Homework Statement

There is a 100kg object on a 30-degree incline ramp.

The mew of friction is 0.5.

There is a motor on top of the ramp, which accelerates the object at .22 m/s^2 by a rope.

Find the power output of the motor.

## Homework Equations

d=(vit)+(.5)(a)(t)^2

W=Fd

F=ma

P=w/t, E/s

## The Attempt at a Solution

I think I'm supposed to use Ek+Ep+TE=0,

but how am I supposed to find the power without knowing the time?

So, I've tried this method:

P=J/s

J(work)= Fxd

So I thought maybe I'll use t=1:

vi=0

vf=x

a= 0.22

d=?

t=1

d=vit+(.5)(a)(t)^2

d=(.5)(.22)(1)

d=.11m

Fnet=ma

Fa-Ff=ma

Fa-(100kg)(9.81)(cos30)(.5[mew])-(100kg)(9.81)(sin30)=(100)(.22)

Fa-424.79-490.5=22

Fa=937.29N

W=Fd

W=(937.29)(.11)

W=103.10J

P=J/s

P=103.10Watts

So here's the one approach I had,

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Here is the second

Ek+Ep+TE=0

(.5)(m)(vf^2-vi^2) + mgh + (Ff)d=0

To find Vf,

Vi=0

Vf=?

a=0.22

d=x

t=1s

a=(Vf-Vi)/t

.22=Vf/1

Vf=.22 m/s

plugging it back in,

(.5)(100kg)(0.0484) + (100)(9.81)(h) + (100)(9.81)(cos30)(0.5[mew])d=0

To find d,

Vi=0

Vf=x

a=.22

d=?

t=1s

d=vit+(.5)(a)(t^2)

done above,

d=0.11m

(.5)(100kg)(0.0484) + (100)(9.81)(h) + (100)(9.81)(cos30)(0.5)(0.11) = 0

2.42 + 981h + 46.73 = 0

h=.05m

Then plug the height of .05m,

to get 98.30J/s?

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Another question:

If there is an elevator moving downwards to the ground at a constant velocity,

Is the work done 0J? Is the net work between Ft and Fg the same? How should I approach this question? I know there is NOT an acceleration. W=Fd, W=mad, so no work?

Although there is a ΔEnergy?

Another question:

Momentum question,

Is the "internal work" done by an explosion equal to the combined kinetic energy of the particles? Or is it 0J? How can i visualize internal work in a bomb-like scenario?

Thank you very much.