Question

31. A vehicle of mass 1250 kg is driven with an acceleration 0.25 ms-2 along a straight level road against an external resistive force 500N, Calculate the power delivered by the vehicle's engine if the velocity of the
vehicle is 30m/s​

Answer 1

case vhole=√(vputt2-2gh). However, there is friction which will slow the ball further and the frictional force on a slope (f=μmgcosθ) will be different than on level ground (f=μmg); here μ is the coefficient of friction (essentially, I believe, what is measured by the "stimp meter"), m is the mass of the ball, and g is the acceleration due to gravity. Since the frictional force is smaller on the slope, you might think that less energy is lost to friction on the slope. But, what matters is not the force but its product with the distance s over which it acts, specifically the work done by friction is Wf=-fs where the negative sign indicates that the friction takes energy away from the ball. When the ball is moving forward along along the segment L2, the distance it travels is s=L2/cosθ and so the work done is Wf=fs=(μmgcosθ)(L2/cosθ)=μmgL2, exactly the same as if the slope was not there. The total work done by friction, regardless of the path, is Wf=μmgL. The final velocity is then vhole=√[vputt2-2g(h+μL)].

Answer 2

Answer:

The vehicle’s engine has to do work against

resistive force and make vechile to move

with an acceleration. Therefore, power

delivered by the vehicle engine is

Explanation: