Energy/Work Question

1. The problem statement, all variables and given/known data

A mass of .263 kg is dropped on a spring with k=252 N/m. When it lands, it compresses the spring 11.8 cm. a) While the spring is being compressed, how much work is being done by gravity? b) What is the work done by the spring as it is being compressed? c) What is the speed of the block when it hits the spring?

2. Relevant equations

W=Fdcos theta
W= -.5kx^2
deltaEmechanical=0


3. The attempt at a solution

I got part a by using -(.263)(9.8)(.118)=-.304 J
I got part b by using W=-.5 (252)(-.118)^2=-1.75 J

I have a question for the third part.

I set up

(Kf+Pf)=(Ki +Pi)
(0+ .5kx^2)=(.5mv^2+ Pi)

I thought that Pi would be zero since it's all KE by the time it hits the spring, but my teacher solved the problem using the negative of the work done by gravity when the spring was being compressed. I am confused on why you use .304 J for Pi. Please explain.
 

Doc Al

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I got part a by using -(.263)(9.8)(.118)=-.304 J
I got part b by using W=-.5 (252)(-.118)^2=-1.75 J
Why the minus sign for part a?

I have a question for the third part.

I set up

(Kf+Pf)=(Ki +Pi)
(0+ .5kx^2)=(.5mv^2+ Pi)
This is a bit confusing since you don't specify spring PE versus gravitational PE. I would use:
Ki + gPEi + sPEi = Kf + gPEf + sPEf

I thought that Pi would be zero since it's all KE by the time it hits the spring, but my teacher solved the problem using the negative of the work done by gravity when the spring was being compressed. I am confused on why you use .304 J for Pi. Please explain.
The work done by gravity is positive, since both force (gravity) and displacement are in the same direction.

Another approach is to simply say:
KEf = KEi + work done.

Since you already figure out the work done in parts a and b, why not use it.
 

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