# Work Energy Concept

1. Jan 27, 2007

### fffff

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

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

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.

Last edited: Jan 27, 2007
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