# Mechanical Power, Physics Homework Help - Sample Exercises Fully Solved

Solved Physics Homework Problems on Mechanical Power.

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Example One of our Homework Help, Solved Problem on Mechanical Power Using Formula Force Times Velocity.

A car weighting 1920 lb and is moving on a horizontal path reaching a maximum velocity of 96 ft/s when the engine runs at its maximum power, 48 hp. Calculate the maximum velocity of the car when it is going up a hill with a slope of 8%. Assume that the air resistance is constant.

Solution to this solved physics homework example:

The hill forms an angle with the horizontal equal to
arctg0.08 = 4.57º.

Transforming the units to the British System,
48 hp = 48*550 ft lb/s = 26.400 ft lb/s = max. power.

First, we need to calculate the air resistance force.

Fengine - Fair= mass*accel. = 0. As the acceleration is zero, Fengine = Fair.

Using Power = Force*velocity,
Fengine = Power/velocity = (26400 ft lb/s)/(96 ft/s) = 275 lb.

This is also the air resistance force, Fair = 275 lb.

The motion of the car along the hill is due to the force F (=Power/velocity) exerted by the engine minus the force 1920*sin4.57º = 153 lb due to the weight of the car and minus the force of 275 lb due to the air resistance. As the acceleration is zero, we can write:

F - 153 lb - 275 lb = mass*accel. = 0 or F = 428 lb.

For maximum velocity, we replace F by (max. power)/v. (26.400 ft lb/s)/v = 428 lb. Solving for v results v = 61.7 ft/s.

Example Two of our Mechanical Power Homework Help. Velocity required When the Resistance Force is Proportional to the Velocity.

Without pedalling, a cyclist goes down a road inclined 5º with a 6 m/s constant speed. The total mass for the cyclist and the bicycle is 70 kg. a) What should it be the power developed by this cyclist to go up the same road at the same speed.
b) Assuming the resistance force Fr is proportional to the speed v, that is Fr = kv, and a 1800 W maximum power done by the cyclist, calculate the maximum speed he can reach going up the road.

Solution:

a) As the cyclist goes down with a constant speed, his acceleration is zero. Denoting Fr as the resistance force, then mgsin5º - Fr = 0 Or Fr = mgsin5º = 59.8 N.

When going up with the same speed, Fr is the same, Fr = 59.8 N. Denoting Fc as the cyclist force, we write
Fc - Fr - mgsin5º = 0 Or Fc = 2Fr = 119.6 N.

Then cyclist Power = Fc * velocity = 119.6 N * 6 m/s = 717.6 Watts.

b) Fr = kv.
k is calculated knowing that when v = 6 m/s, Fr = 59.8 N.
k = 59.8/6 kg/s = 9.97 kg/s.
Fr = 9.97kg/s*v.

Again, Fc - Fr - mgsin5º = 0 and Power = Fc * velocity v . As the maximum speed corresponds to a maximum power, Fc = 1800 W/v. Replacing and disregarding the units and knowing that v will be in m/s:

1800/v - 9.97v - 70*9.8*sin5º = 0

Or, 1800 - 9.97v2 - 59.8v = 0. Solving this quadratic equation results in v = 10.8 m/s

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