You are Here: Home >< Physics

Kinetic energy watch

1. A van is pushed from rest with a force of 130N over a distance of 60m. The resistivity forces acting on the van total 90N

Using the information above how would i calculate the kinetic energy the van gains?
2. Are you given the mass of the van?
3. (Original post by JBKProductions)
A van is pushed from rest with a force of 130N over a distance of 60m. The resistivity forces acting on the van total 90N

Using the information above how would i calculate the kinetic energy the van gains?
Work done = force x distance. So if you can work out the total work done by the van over the distance and subtract the work done against resistive forces, you will have the increase in kinetic energy
4. (Original post by D-Day)
Are you given the mass of the van?
Nope thats all ive been given
5. (Original post by sonofdot)
Work done = force x distance. So if you can work out the total work done by the van over the distance and subtract the work done against resistive forces, you will have the increase in kinetic energy
Or that.
6. (Original post by sonofdot)
Work done = force x distance. So if you can work out the total work done by the van over the distance and subtract the work done against resistive forces, you will have the increase in kinetic energy
so is it 2400 J?
7. 2400 J pr 2.4 Kj
8. (Original post by JBKProductions)
so is it 2400 J?
Indeed
9. (Original post by jjkkll)
2400 J pr 2.4 Kj
ok thanks
10. (Original post by sonofdot)
Indeed
if you dont mind me asking another question. The resistance to the motion of a motor boat of mass 780kg is 120N, and the driving force from the propellers is 250N. What distance does it cover while increasing from 2 to 3m/s??
11. (Original post by JBKProductions)
if you dont mind me asking another question. The resistance to the motion of a motor boat of mass 780kg is 120N, and the driving force from the propellers is 250N. What distance does it cover while increasing from 2 to 3m/s??
I would help, but somebody negged me inexplicably on this thread...

I'll help really This is like the opposite way around. Find the increase in the kinetic energy () and then the total work done equals the kinetic energy plus the work done against resistance
12. (Original post by sonofdot)
I would help, but somebody negged me inexplicably on this thread...

I'll help really This is like the opposite way around. Find the increase in the kinetic energy () and then the total work done equals the kinetic energy plus the work done against resistance
how would i calculate the kinetic energy without the velocity or the distance it travels?
13. (Original post by JBKProductions)
how would i calculate the kinetic energy without the velocity or the distance it travels?
you have the velocity. The gain in kinteic energy = kinetic energy at 3m/s - kinetic energy at 2m/s
14. (Original post by sonofdot)
you have the velocity. The gain in kinteic energy = kinetic energy at 3m/s - kinetic energy at 2m/s
oh ok i see now i didnt understand the question properly i thought i had to work out acc and etc lol but ill try and come bk if i get stuck lol
15. (Original post by sonofdot)
you have the velocity. The gain in kinteic energy = kinetic energy at 3m/s - kinetic energy at 2m/s
so the kinetic energy is 1950J?
16. (Original post by sonofdot)
I would help, but somebody negged me inexplicably on this thread...

I was not me you crack head, at least ask before negging lol.
17. (Original post by jjkkll)
I was not me you crack head, at least ask before negging lol.
Ok sorry I guess I miscalculated your power... not like my rep is worth anything

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 14, 2008
Today on TSR

Should I drop out of uni

...to become a pro gamer?

University open days

• University of Buckingham
Fri, 14 Dec '18
• University of Lincoln
Mini Open Day at the Brayford Campus Undergraduate
Wed, 19 Dec '18
• University of East Anglia