A 1400 kg car is accelerating up a hill. The hill is 150 m long and the total rise of the hill is 6.0 m. The car accelerates from a speed of 7.0 m/s at the bottom to 15 m/s at the top, in 12 s. ...
A 1400 kg car is accelerating up a hill. The hill is 150 m long and the total rise of the hill is 6.0 m. The car accelerates from a speed of 7.0 m/s
at the bottom to 15 m/s at the top, in 12 s. If the average retarding force of friction is 700 N, find the average power of the car.
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This can be done by the conservation of energy.
At the bottom of the hill car only has its kinetic energy. At the bottom car power up for 12s and reach the hill top with a velocity of 15m/s. At this process the car uses its power to overcome the friction force and the component of it's own weight that act against its direction of travel.
P = average power of car
Let `theta` be the slope angle of the hill with horizontal.
`sintheta = 6/150 = 1/25`
Weight component act against cars' travel `= 1400g*sintheta = 56g`
Let the bottom horizontal be the datum.
Kinetic energy at bottom `= 1/2*1400*7^2 = 34300J`
Kinetic Energy at top `= 1/2*1400*15^2 = 157500J`
Potential energy at bottom `= 0`
Potential energy at top `= 1400*9.81*6 = 82404J`
Work done against friction `= 700*150 = 105000J`
Work done against own weight `= 56*9.81*150 = 82404`
By law of energy conservation;
Energy at bottom + work done by car = energy at top+work done against friction and its own weight
`0+34300+P*12 = 157500+82404+105000+82404`
` P = 32.75 (KJ)/s`
So the average power of car is 32.75 KW.
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