Download The Law of Conservation of Mechanical Energy

Henry Ford Community College
Science Division
Winter 2014
Physics 131, Section 70
Conservation of Mechanical Energy, Power and Momentum
The Law of Conservation of Mechanical Energy
The law of conservation of mechanical energy states that “The sum of the kinetic energy and
the potential energy in a system is constant if no resistant forces do work”.
Mathematically, this law can be stated as:
KE1 + PE1 = KE2 + PE2 = …….KEn + PEn
In other words the sum of the kinetic and potential energy at state 1 is equal to the sum of the
kinetic energy and potential energy at the nth state.
Questions on The Law of Conservation of Mechanical Energy
Question 1 (To be solved in class)
A 10 kg mass is dropped from a hot air balloon at a height of 350 m above the ground. Find its
speed at points:
a.
b.
c.
d.
300 m above the ground
200 m above the ground
100 m above the ground
As it hits the ground.
Question 2 (To be solved in class)
An ironworker working on a building at a height of 10 m above the ground drops a hammer.
What is the velocity of the hammer when it is 4 m above the ground?
Question 3 (Student Self Practice)
A skydiver jumps from a hovering helicopter that is 3000 m above the ground. If air resistance is
ignored,
a. How fast will he be falling when his altitude is 2000 m ?
Power
Power is defined as the rate of doing work (how fast work is done).
Mathematically,
Power = Work/Time
Work is measured in Joules (J)
Time is measured in seconds (s)
Power is measured in Watt (W)
Questions on Power
Question 1 (To be solved in class)
A freight elevator with operator weighs 5000 N. If it is raised to a height of 15 m in 10 s, how
much power is developed.
Question 2 (To be solved in class)
A 525 kg steel beam is raised 30 m in 25 seconds. What is the power needed?
Question 3 (To be solved in class)
A motor on an escalator is capable of developing 12 kW of power.
a. How many passengers of mass 75 kg each can it lift a vertical distance of 9 m per minute
assuming no loss of power?
b. What power from the motor is needed to move the same number of passengers at the
same rate if 45% of the actual power developed by the motor is lost to friction and heat
loss?
Question 4 (To be solved in class)
An electric motor develops 65 kW of power as it lifts a loaded elevator 17.5 m in 35 s. How
much force does the motor exert?
Question 5 (Student Self Practice)
A boy whose mass is 50 kg finds that he can run up a flight of 40 steps each 16 cm high in 5.2 s.
What is the power developed by the boy as he performs this exercise?
Question 6 (Student Self Practice)
What power does a pump develop to lift 35 L of water per minute from a depth of 110 m? (A
liter of water has a mass of 1 kg)
Momentum
Definition of Momentum
Momentum is a measure of the difficulty in bringing a moving body to rest.
It is a vector quantity whose direction is that of the velocity.
Mathematically, momentum, denoted by p is given by
p = m.v
m is mass in kg
v is velocity in m/s
p is momentum. The unit for momentum is kg m/s
Newton’s second law of motion stems from the fact that the force applied on a body is directly
proportional to the time rate of change of momentum.
Mathematically,
F = (mvf – mvi)/t
F = m(vf – vi)/t
F = ma
F is applied force in N
m is mass in kg
vi is initial velocity in m/s
vf is final velocity in m/s
t is time in s
Impulse
Impulse causes a change in momentum.
Thus, the change in momentum produced by an impulse is equal to the impulse in both
magnitude and direction.
Impulse = Change in Momentum
From Newton’s second law,
F.t = mvf = mvi
Thus impulse has units of kg m/s OR Ns
Questions on Momentum and Impulse
Question 1 (To be solved in class)
a. Find the momentum of an automobile of mass 1350 kg travelling at 20 m/s
b. Find the momentum of an automobile of mass 1350 kg travelling at 65 km/h
c. A vehicle travelling at 30 m/s possesses 32,000 kg m/s of momentum. Find the mass of
the vehicle.
Question 2 (To be solved in class)
Find the velocity that a bullet of mass 1.00 X 10-2 kg would have to have so that it has the same
momentum as a lighter bullet of mass 1.8 X 10-3 kg and velocity 325 m/s.
Question 3 (To be solved in class)
A golfer strikes a golf ball of mass 0.05 kg and the time of impact between the golf club and the
ball is 1 millisecond. If the ball acquires a velocity of magnitude 75 m/s, calculate the average
force exerted on the ball.
Question 4 (To be solved in class)
A football team’s kicker punts the ball (mass = 0.42 kg) and gives it a launch speed of 30 m/s.
Find the impulse delivered to the football by the kicker’s foot and the average force exerted by
the kicker on the ball, given the impact time is 8 milliseconds.
Question 5 (Student Self Practice)
A sedan of mass 1500 kg is moving at 100 km/h
a. Find its momentum
b. At what velocity is the momentum of a larger car of mass 2200 kg equal to that of the
smaller car?
Question 6 (Student Self Practice)
A snowmobile has a mass of 240 kg. A constant force is exerted on it for 55 seconds. The
snowmobile’s initial velocity is 5 m/s and its final velocity is 25 m/s.
a. What is the change in momentum?
b. What is the force exerted on it?
Question 7 (Student Self Practice)
A 0.144 kg baseball is pitched horizontally at 38 m/s. After it is hit by the bat, it moves at the
same speed, but in the opposite direction.
a. What was the change in momentum of the ball?
b. What was the impulse delivered by the bat?
c. If the bat and the ball were in contact for 1 millisecond, what was the average force the
bat exerted on the ball?