Work & Energy (Question on Classical Mechanics/Slope based Problems)

• warhammer
In summary, the conversation discusses the calculation of work done and energy expended using the Change in Kinetic Energy and Work Energy Theorem. It is noted that there are two different functions of work done, one by the Gravitational Force and one by frictional force or brakes. The calculations are used to determine if the car is going uphill or downhill, with a mistake initially made in the calculation being pointed out and corrected. The conversation concludes with the acknowledgement of an error and discrepancy in the book's solution.

warhammer

Homework Statement
A car weighing 1350 kg is going down a hill. When it is 60 m vertically above the bottom of the hill, the driver sees red light of traffic crossing at the bottom. His speed at the time brakes are applied is 20 ms. How much energy will be dissipated by the brakes if wind and other frictional effects are neglected. Take g = 9.80 ms ².
Relevant Equations
Work Energy Theorem: /Delta Kinetic Energy = Work Done
I used the Change in Kinetic Energy and equated that with the Work Done. The "Work Done" part comprises of two different functions- one is work done by Gravitational Force while the other is the work done by frictional force (or the brakes).

/Delta KE (magnitude wise)= 0.5*1350* (20^2)=270,000 J=270 kJ ---(1)

Work Done by Gravity= 1350*9.8*60=793,800---(2)

Work done by Friction=W---(3)

Adding (2) & (3) & equating with (1) we get W=793800-270000=523,800 J

I used the idea here that at said height the car has both KE and PE, thus I used both in the Work Energy Theorem to calculate work done/energy expended.

Is the car going uphill?

Careful with signs. ΔKE = net work done. Note that ΔKE is negative while the work done by gravity is positive.

warhammer
PeroK said:
Is the car going uphill?
No sir it is going downhill

warhammer said:
No sir it is going downhill

warhammer and Doc Al
Doc Al said:
Careful with signs. ΔKE = net work done. Note that ΔKE is negative while the work done by gravity is positive.
Thank you pointing out that mistake sir.

As ΔKE=-270,000, then the W= -270,000-793,800=−1063800.

Now you've got it.

warhammer
PeroK said: