# Work Energy Concept

1. Jan 27, 2007

### fffff

1. The problem statement, all variables and given/known data
http://xs511.xs.to/xs511/07046/workenergy.JPG [Broken]

2. Relevant equations

3. The attempt at a solution

Ok im all confused on this one.

ok so leave part (I) and (II)

onto part (III)

word done by the pulling force acting on the block = the loss of energy + work done by resisting force

by loss of energy i mean gain in PE - loss in KE

=(7500+225)-(1225)+(7.5 x 200) = 8000J

now im confused as to whether or not the value of theta is mgsintheta or fscostheta please help

Last edited by a moderator: May 2, 2017
2. Jan 27, 2007

Don't overcomplicate things. The work done by all the forces acting of the block from A to B equals the change in kinetic energy from A to B. Regarding the trig, draw a sketch.

3. Jan 27, 2007

### fffff

ok then the loss in KE=1000J
but thats not the right answer, theres also resistance forces how do i take that into account and what do you mean by internal and external forces.

http://xs511.xs.to/xs511/07046/trainagent.JPG [Broken]

Last edited by a moderator: May 2, 2017
4. Jan 27, 2007

work done by pulling force + work done by resistance force = change in kinetic energy. => (iii) work done by pulling force = change in kinetic energy - work done by resistance force. This should do.

5. Jan 27, 2007

### fffff

work done by pulling force = 1000 -(7.5x200)=-500 J

i though it was

work done by pulling force= gain in ke + gain in ke + work done by resisting force

but if its loss in ke then we minus it

6. Jan 27, 2007

### fffff

just bump this up.

7. Jan 27, 2007

### arildno

Let's take one question at a time:

How would you calculate the answer to i)?

8. Jan 28, 2007

### fffff

green pandas

loss in KE=1/2m(v^2-u^2)=
1/2(50)(49-9)=1000J

Gain in PE=mgh=(50)(10)(15)=7500J

so has the car goes up the straight hill it gains 7500J in PE but looses 1000J KE.
Because thers a resistance force of 7.5 N, the work done by it=fs=7.5(200)=1500J

So work done py pulling force=(7500-1000)+(1500)=8000J

9. Jan 28, 2007

### fffff

thats wrong

work done by pulling force - work done by resistance=change in kinetic energy

10. Jan 28, 2007