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Mechanical Energy and Conservation
Introduction
Problems become very simple if only conservative forces are
involved.
/\KE + /\PE = 0
E = Total Mechanical Energy = KE + PE The
total mechanical energy remains constant if only conservative forces are
used. This is the principle of conservation of mechanical energy. In
science when we say energy is conserved, we mean that the total amount of energy
remains constant. This is different from the everyday use of the term
conserved. Example we conserve fuel by using a fuel efficient car. 
The mass starts off at h. It is motionless, thus its
Kinetic Energy is 0.
When the mass has
dropped 1/2 h, its Potential Energy is cut in 1/2 and converted into Kinetic
Energy. When
at 0 h, the Potential Energy is 0 and all the energy is Kinetic Energy. Table of Contents Problems
| 1. |
A 60.0 kg block of ice is at the top of an
inclined plane with a height of 10.0 m. (a) What is the velocity of
the block of ice at the bottom of the hill? (b) What is the height
at 1/2 the velocity at the bottom of the hill?
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| 2. |
Calculate the kinetic energy and velocity
required for a 80.0 kg pole vaulter to pass over a bar 6.00 m above the
ground. Assume the vaulter's center of mass is initially 0.90 m
above the ground and reaches its maximum height at the level of the bar.
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| 3. |
A 6.00 kg bowling ball is pressed against a
horizontal spring in a tube. The spring with a constant of 350. N/m
is compressed 45.0 cm. The tube is on a cliff 10.0 m above the
ground. How far from the base of the cliff does the bowling ball hit
the ground?
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| 4. |
A 1500. kg car is dropped 5.00 m and the
springs of the car are compressed 35.0 cm when the car hits the
ground. What is the spring constant of each of the four springs in
the car? |
Table of Contents Answers
| 1. |
A 60.0 kg block of ice is at the top of an
inclined plane with a height of 10.0 m. (a) What is the velocity of
the block of ice at the bottom of the hill? (b) What is the height
at 1/2 the velocity at the bottom of the hill?
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(a) All of the Potential Energy at the top of the
hill is converted into Kinetic Energy at the bottom of the hill.
mgh = 
mv2
Solve for velocity.
v = ( 2gh ) 1/2 = ( 2 x
9.80 m s -2 x 10.0 m ) 1/2 = 14.0 m/s
(b) Total Energy = PE + KE
At the top of the hill, all of the energy is Potential Energy.
Total Energy = mgh
mgh = mgh' +
mv 2
Velocity is 7.00 m/s exactly 1/2 the velocity
at the bottom of the hill.
gh -
v 2
9.80 m s -2 x 10.0 m -
( 7.00 m/s ) 2
h' = ------------------- =
--------------------------------------------------- = 7.50
m
g
9.80 m s -2
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| 2. |
Calculate the kinetic energy and velocity
required for a 80.0 kg pole vaulter to pass over a bar 6.00 m above the
ground. Assume the vaulter's center of mass is initially 0.90 m
above the ground and reaches its maximum height at the level of the bar.
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The pole vaulter must have sufficient
Kinetic Energy to convert into Potential Energy for the difference in the
center of height at the beginning to the height at the level of the bar.
/\h = 6.00 m - 0.90 m = 5.10 m
mg/\h =
mv 2
Solve for v.
v = ( 2g/\h ) 1/2 = ( 2 x 9.80 m s -2
x 5.10 m ) 1/2 = 10.0 m/s
KE = mv 2 =
x 80.0 kg x (10.0 m/s) 2
= 4.00 x 10 3 J
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| 3. |
A 6.00 kg bowling ball is pressed against a
horizontal spring in a tube. The spring with a constant of 350. N/m
is compressed 45.0 cm. The tube is on a cliff 10.0 m above the
ground. How far from the base of the cliff does the bowling ball hit
the ground?
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W = KE
kx 2 =
mv
2
Solve for v.
kx 2
350 N m -1 x ( 0.450 m ) 2
v = ( ------- ) 1/2 = (
------------------------------------ ) 1/2 = 3.44
m/s
m
6.00 kg
Use the height and the acceleration of gravity
to calculate the time the bowling ball is in the air. Because the
spring is applying a horizontal force on the ball, the initial velocity of
the ball in the vertical direction is 0.
x =
g t2
Solve for t.
2x
2 x 10.0 m
t = ( ----- ) 1/2 = (
----------------- ) 1/2 = 1.43 s
g
9.80 m s -2
Use the time the bowling ball is in the air to
calculate the horizontal displacement using the horizontal velocity.
Remember, there is no acceleration in the horizontal direction.
x = vh t = 3.44 m/s x 1.43 s = 4.92
m
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| 4. |
A 1500. kg car is dropped 5.00 m and the
springs of the car are compressed 35.0 cm when the car hits the
ground. What is the spring constant of each of the four springs in
the car?
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The Potential Energy of the falling car is converted
into Kinetic Energy at the instant the car hits the ground. This
does work in compressing the springs of the car. Each spring
converts 1/4 of the Potential Energy.
PE = mgh
W =
kx 2
mgh =
kx 2
Solve for k.
2 mgh
2 x ( 1500. kg / 4) x 9.80 m s -2 x 5.00 m
k = ----------- =
------------------------------------------------------- = 3.00 x 10 5
N m -1
x2
( 0.350 m ) 2
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Table of Contents | 
