# Work Energy Problems

1. Nov 25, 2003

### Az_

Well i am taking physics cp so these problems should be fairly easy but i've been sitting here for about an hour and somehow i am still drawing a blank. I did these problems with ease last week but somehow i am utterly dumbfounded now without the slightest hint on how to do these. I think i am missing information but its all there, Its just so confusing. I am hoping with these 3 questions answered my brain will actually start moving again so i can finish the rest of the 30 problems and start on my Trig and Statistics hw.

A laborer at the pickle boxing plant needs to send a 50kg crate of Kosher-Dills up a 5m long ramp that is inclined at an angle of 30 degrees. The coefficient of friction on the ramp is .25. Use conservation of energy to find the minimum velocity the crate needs to have a the bottom of the ramp in order to make it to the top (assuming it stops right at the top).

I have no idea how to do this one...

One Food calorie is equal to 1000 energy calories and is equal to 4186 J. If during a jumping jack a person lifts his center of mass about .30 meters and he has a mass of 70 kg, how many jumpng jacks does he need to do in order to expend 1 food calorie of energy assuming the human body is 100% efficient.

Well W=FD right? D is .30 and the number of jumping jacks would be .30(x) with x. W would be 4186 J. And all i do is solve for x but i have a hard time remembering how to find F.

A 3 kg mass is hung from a spring, and stretches it 35 cm. What is the spring constant.

Well i know i have to use Ue = .5(k)(.35)^2 but how do i find K if i dont know Ue???

Last edited: Nov 25, 2003
2. Nov 25, 2003

Well W=FD right? D is .30 and the number of jumping jacks would be .30(x) with x. W would be 4186 J. And all i do is solve for x but i have a hard time remembering how to find F.

W = FD is correct. The work done against gravity is equal to the distance moved upward multiplied by the force of gravity acting on the person, i.e. their weight.

W = FD = mgD

3. Nov 25, 2003

### Az_

Alright thanks. I did that problem out and got 22 jumping jacks. And i then read the post about hooke's law where you stated
x = mg/k so i can do .35 = 3(9.8)/k and solve for K? Is this right?

4. Nov 25, 2003

Yeah, you can use F = kx.

The equation E = &frac12;kx2 comes from the fact that W = FD.

But here you have a force that increases as you extend the spring further, but the average force is equal to

(0 + F)/2 = &frac12;F

And since from Hooke's law

F = kx

E = &frac12;(kx)*x = &frac12;kx2

Since F = kx

k = F/x = mg/x

5. Nov 25, 2003

### ShawnD

Quite a tricky question.
The force that the ramp/ground puts on the box (normal force) is given by the formula Fn = Fg*cos(a) the a is usualy a theta but I don't know how to make those. It's cosine because the force from the ramp is at a maximum when the angle is 0 (when picking a trig function, think min/max).
Fn = (50)(9.81)cos(30)
Fn = 424.785N

The force from friction (f) is given by uFn where u is friction coefficient and Fn is normal force.
f = (0.25)(424.785)
f = 106.196N

Since you know the length of the ramp and the angle, you can find the height. sine(a) = opposite/hypotinuse. Maniupate it to find your height.

Since the box does work against friction and the box gains gravitational potential energy, the formula for the energies looks like this:
(1/2)mv^2 = Fd + mgh

m is mass, v is velocity, F is force (friction in this case), d is distance (length of the ramp), g is gravity (9.81), h is height (vertical height which you calculated)

in = out
4186 = (0.3)(70)(9.81)x
solve for x

kx = mg
k(0.35) = (3)(9.81)
solve for k