To calculate the work done on the car, we can use the work-energy principle. The work done on an object is equal to the change in its kinetic energy.
The initial kinetic energy of the car is given by KE1 = (1/2)mv1^2, where m is the mass of the car (1,500 kg) and v1 is the initial velocity (30 m/s).
The final kinetic energy of the car is given by KE2 = (1/2)mv2^2, where v2 is the final velocity (15 m/s).
The work done on the car is W = KE2 - KE1.
W = (1/2)m(v2^2 - v1^2)
= (1/2)(1,500 kg)((15 m/s)^2 - (30 m/s)^2)
= (1/2)(1,500 kg)(225 - 900) m^2/s^2
= (1/2)(1,500 kg)(-675) m^2/s^2
= - 506,250 J
Therefore, the work done on the car from the brakes is -506,250 J.
A 1,500 kg car’s speed changes from 30 m/s to 15 m/s after the brakes are applied. Calculate the work done onto the car from the brakes.(1 point)%0D%0AResponses%0D%0A%0D%0A−168,750 J%0D%0Anegative 168,750 upper J%0D%0A%0D%0A506,250 J%0D%0A506,250 upper J%0D%0A%0D%0A−506,250 J%0D%0Anegative 506,250 upper J%0D%0A%0D%0A168,750 J
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1 answer