Exercises Fully Solved

**Solved Physics Homework Problems on Mechanical
Power.**

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forward you the online answers for your physics homework as shown with the
problem examples shown below. Our online help arrives as email attachments.**

**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.
F**

** 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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