# Work done on a hill

• Oct 27th 2010, 12:47 AM
gbenguse78
Work done on a hill
I have a slight problem. Please comment on the question and solution.

A car of mass 2 Tonnes arrives at a foot of a hill, travelling at a speed of 10m/s and reaches the top of the hill with a speed of 5m/s. The hill is 500m long and rises 1 in 125. If there is a retarding force of 100N, calculate the amount of work done by the engine in getting the car up the hill.

Solution 1: Energy conservation.
I surmised the following:
Initial KE - Work done= PE gained + Final KE (Is this correct?)
Initial KE= 0.5 X 2000 X 100 = 100KJ
Work done=?
PE Gained= 2000 x 10 x 1/125 = 160J
Final KE = .5 X 2000 X 25 = 25KJ

Work Done = (100-160-25)KJ= -85KJ
• Oct 27th 2010, 06:01 AM
CaptainBlack
Quote:

Originally Posted by gbenguse78
I have a slight problem. Please comment on the question and solution.

A car of mass 2 Tonnes arrives at a foot of a hill, travelling at a speed of 10m/s and reaches the top of the hill with a speed of 5m/s. The hill is 500m long and rises 1 in 125. If there is a retarding force of 100N, calculate the amount of work done by the engine in getting the car up the hill.

Solution 1: Energy conservation.
I surmised the following:
Initial KE - Work done= PE gained + Final KE (Is this correct?)
Initial KE= 0.5 X 2000 X 100 = 100KJ
Work done=?
PE Gained= 2000 x 10 x 1/125 = 160J
Final KE = .5 X 2000 X 25 = 25KJ

Work Done = (100-160-25)KJ= -85KJ

The PE gained is $m \times g \times h$ where $h$ is the change in height

The work done by the retarding force is the force times the distance the car moves in the direction of the force.

CB
• Nov 15th 2010, 08:53 PM
gbenguse78
Hi
A very belated reply to all your help. As a husband, worker and A levels student you can call me superman. Apologies for the delay in responding. Thanks once again