Solve Work Energy Problems: Questions & Answers

AI Thread Summary
The discussion revolves around solving work-energy problems in physics, particularly focusing on a crate being moved up a ramp, the energy expended during jumping jacks, and calculating the spring constant of a spring. Participants express confusion over applying conservation of energy principles and calculating forces, with specific examples provided for clarity. Key formulas discussed include W = FD for work done against gravity and the relationship between force, mass, and acceleration due to gravity. The conversation emphasizes the importance of understanding how to manipulate these equations to solve for unknowns in various scenarios. Overall, the thread serves as a collaborative effort to clarify fundamental physics concepts related to work and energy.
Az_
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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.
Thnks in Advance.

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 don't know Ue?
 
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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
 
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?
 
Yeah, you can use F = kx.

The equation E = ½kx2 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 = ½F

And since from Hooke's law

F = kx

E = ½(kx)*x = ½kx2

Since F = kx

k = F/x = mg/x
 
Originally posted by Az_
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).


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)


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


A 3 kg mass is hung from a spring, and stretches it 35 cm. What is the spring constant.
kx = mg
k(0.35) = (3)(9.81)
solve for k
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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