Someone is driving down a slope of 8 degrees to the horizontal, they then apply the brakes to remain at a constant speed of 60km/h. The breaks pads of the car weigh 100grams each and there are two on each wheel. They have a heat capacity of 4 J/K per gram and they can tolerate temperatures of up to 300 degrees C.
How long (in seconds) will it take for the breaks to reach 300 degrees c?
How long (in seconds) will it take for the breaks to reach 300 degrees c?
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You didnt mention initial temperature of the break pads and the mass of the car!!!
Ok,lets consider it is 273K(So that it will easier to calculate)
Now the break pads,can tolerate temp upto 300C(573K)
Energy required to raise temp. of each pad to 300C=4*100*300=120kJ
Therefore,for 8 pads(2 in each wheel),Energy required=960kJ
Now,since the car has constant speed,
Weight of the car*sin 8(Degrees)=Frictional Force by Brake pads.Also,Work done by Gravity =Work done by break pads
Hence,
Weight*sin 8(degrees)*Distance travelled=960kJ
=>Distance=960kJ/Weight*sin 8(deg)
=>Time(in hrs)=960/W*sin 8*60
=160/Wsin8
Time (in sec)=3600*Time(in hours)
P.S.-You can also convert speed to m/s and do this.Hope,I helped.
Ok,lets consider it is 273K(So that it will easier to calculate)
Now the break pads,can tolerate temp upto 300C(573K)
Energy required to raise temp. of each pad to 300C=4*100*300=120kJ
Therefore,for 8 pads(2 in each wheel),Energy required=960kJ
Now,since the car has constant speed,
Weight of the car*sin 8(Degrees)=Frictional Force by Brake pads.Also,Work done by Gravity =Work done by break pads
Hence,
Weight*sin 8(degrees)*Distance travelled=960kJ
=>Distance=960kJ/Weight*sin 8(deg)
=>Time(in hrs)=960/W*sin 8*60
=160/Wsin8
Time (in sec)=3600*Time(in hours)
P.S.-You can also convert speed to m/s and do this.Hope,I helped.
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Thanks!
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