Download Conservation of Energy and Power

Conservation of Energy and Power
1. An electrical motor lifts a 575 N box 20 m straight up by a rope in 10 s. What power is
developed by the motor?
F = 575 N the amount of force the rope must apply to the object to lift it up at constant velocity is
equal to the object’s weight
W
F = 575 N
P=
t
d = 20 m
W =Fd
Fd
t
(575 N)(20 m)
P=
= 1.15×103 W
10s
t = 10 s
P=
P=?
2. A block slides down a frictionless inclined plane of height h = 1 m, making angle θ with the
horizontal. At the bottom of the plane, the block continues to move on a flat surface with a
coefficient of friction µ = 0.30. How far does the mass move on the flat surface?
We apply the law of Conservation of energy. The equation is
ΔE = ΔK + ΔU
The velocity of the block is zero both in the beginning and the end. Thus, ΔK = 0
We also have
ΔE = W friction = − µk m g d
The change in GPE is
ΔU = −mgh
Putting it all together
ΔE = ΔK + ΔU
− µk m g d = −mgh
d=
h
µk
=
1m
= 3.3m
0.30
3. A cyclist approaches the bottom of a hill at a speed of 11 m/s. The hill is 6 m high. Ignoring
friction, how fast is the cyclist moving at the top of the hill? Assume that he doesn’t peddle and
ignore air resistance.
Since there is no friction, the total mechanical energy is conserved. Thus,
ΔU + ΔK = 0 .
ΔU = mgh − 0 = mgh and ΔK = 12 mv2f − 12 mvi2 ,
Substitute these into the energy conservation equation
mgh + 12 mv 2f − 12 mvi2 = 0
v 2f = vi2 − 2 gh
Thus, v f = vi2 − 2 gh = (11m/s)2 − 2(9.81m/s2 )(6m) =
3.28m 2 /s 2 = 1.81m/s .
4. A 100 kg mass traveling with a velocity of 15 m/s on a horizontal surface strikes a spring with a
spring constant k = 5 N/m.
a. Find the compression of the spring required to stop the mass if the surface is frictionless.
Since there is no dissipative force, we will use the Law of Conservation of Energy.
ΔU + ΔK = 0 .
ΔU = 12 kx2 − 0 = 12 kx2
ΔK = 0 − 12 mv2 = − 12 mv2
1
2
kx2 − 12 mv2 = 0
Solve for x:
m
v
k
100 kg
x=
(15 m/s) = 67.1m
5 N/m
x=
b. Find the compression of the spring if the surface is rough ( µk = 0.4 0).
The more general form of the equation including nonconservative forces is
ΔE = ΔU + ΔK
r r
ΔE = F • d = −µk m g x
ΔU = 12 kx2 and ΔK = − 12 mv2
Putting it all together we get:
−µk m g x = 12 kx2 − 12 mv2
where
− µk m g x = −0.40(100 kg)(9.81 m/s 2 ) x = (−392.4 N) x
and
1
2
kx2 = 12 (5N/m) x2 = (2.5N/m) x2
and
1
2
mv2 = 12 (100kg)(15m/s)2 = 11250N
Simplifying (and leaving off the units to make the quadratic equation easier to read)
−392.4 x = 2.5 x 2 − 11250
In standard form this quadratic equation becomes
2.5 x 2 + 392.4 x − 11250 = 0
Solving by the quadratic formula we obtain
x = 24.8 m or x = − 182 m
The physical solution here is positive. Thus, x = 25 m.
5. An amusement park roller coaster car has a mass of 250 kg. During the ride, it is towed to the
top of a 30 m hill, where it is released from rest and allowed to roll. The car plunges down the hill,
then up a 10 m hill and through a loop with a radius of 10 m. Assume that the tracks are
2
frictionless. (Use g = 10 m/s .)
a. What is the Potential Energy of the car at the top of the 30 m hill?
GPETop of Hill = m g h
=
( 250 kg ) (10 m/s 2 ) (30 m )
= 75000 J
b. What are the Kinetic Energy and the speed of the car at the bottom of the 30 m hill?
GPETop of Hill = ETot = KEBottom of Hill = 75000 J
1
m v2
2
2 KE
2 (75000 J)
=
=
m
250 kg
KEBottom of Hill =
v =
600 m 2 /s 2 = 24 m/s
c. What are the Kinetic Energy and the speed of the car at the top of the 10 m hill?
( 250 kg ) (10
GPE10 m = m g h =
m/s 2 ) (10 m ) = 25000 J
GPE10 m + KE10 m = ETot = 75000 J
KE10 m = ETot − GPE10 m = 75000 J − 25000 J = 50000 J
v10 m =
2 KE
=
m
2 (50000 J)
=
250 kg
400 m 2 /s 2 = 20 m/s
o
d. If the hill makes an angle of 60 with the horizontal and the car takes 15 seconds to be towed
up the hill, determine the length of the hill, the velocity of the car, the force required to tow the car
up the hill, and the power of the motor pulling the car up the hill.
30 m = L(sin 60o )
30 m
= 34.64 m
sin 60o
F|| = Wt ⋅ sin 60o
L=
(2500 N)( sin 60o ) = 2165 N = 2200 N
v=
L 34.64 m
=
= 2.31m/s = 2.3m/s
Δt
15s
P=
Work Done 75000 J
=
= 5000 J/s = 5000 W
Δt
15s
or
P = F v = (m g)(sin 60o )(v)=(250 kg)(10 m/s 2 )(sin 60o )(2.31 m/s)
= (2165 N) ( 2.31m/s ) = 5001J/s = 5.0 kW