# Work and Power Car problem

• ludakrishna

#### ludakrishna

1.) A 1900-kg car experiences a combined force of air resistance and
friction that has the same magnitude whether the car goes up or down a
hill at 27 m/s. Going up a hill, the car's engine needs to produce 47 hp
more power to sustain the constant velocity than it does going down the
same hill. At what angle is the hill inclined above the horizontal?

P=F*v*cos (thetha)

I am completely lost. Any help will be necessary. I just need a beginning. Anyways, here's my attemp. mg sign (thetha). 1900(9.8) sin (thetha) and solve for thetha to get the answer. But for some reason i know that i am wrong.

Draw free body diagrams of the car going up and down the hill.

The forces are the weight of the car, the drag and friction force (the same magnitude in both cases) and the force from the engine.

You don't know the drag and friction force so call it F. Find the engine force going up and down the hill, then find the engine power going up and down.

You know Power(up) = Power(down) + 47hp.

The force required to push a mass m up an incline is equal to

F (up) = (mgsinθ - friction )

When going down power is required only to overcome friction

F (down ) = ( friction )

Power = Force x velocity

P (up) –P(down) = ( force up – force down ) x velocity = 47 hp

Therefore extra power for going uphill (47 hp) is equal to mgsinθ x velocity.

I Substituted and fond out the value of theta.

1900*9.81 sin thetha * 27 = 47*746

I calculated theta as ~ 4 degrees.

The force required to push a mass m up an incline is equal to

F (up) = (mgsinθ - friction )

F (up) = (mgsinθ + friction )

F (up) = (mgsinθ + friction )
Better Watch your units on the rest of the question!

When going down power is required only to overcome friction

F (down ) = ( friction )[\QUOTE]

No. Going down the mg sinθ force is still in the equation, but the sign is reversed (the force is trying to increase the speed of the car not decrease it)

F(up) = (friction) + mgsinθ
F(down) = (friction) - mgsinθ