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

[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.

 

[PeroK]

Is the car going uphill?

 

[Doc Al]

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

 

[warhammer]

Is the car going uphill?

No sir it is going downhill

 

[PeroK]

No sir it is going downhill

Your calculations suggested otherwise!

 

[warhammer]

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.

 

[Doc Al]

Now you've got it.

 

[warhammer]

Your calculations suggested otherwise!

Yes that was a grievous error on my part sir!
I was also thrown into a catch because of the solution of the book which neither matched my value nor the sign, so yes