Download vi - Lakota East High School

Energy Practice Problems
Name _______________________________________________ Date ___________________
1.
A 3.00 kg crate slides down a ramp at a loading dock. The ramp is 1.00 m in length and inclined at a angle of
30.0º. The crate starts from rest at the top. Neglecting any resistive forces, use energy methods to determine
the speed of the crate when it reaches the bottom of the ramp.
h
1meter
h = 0.500meter
Ui + Ki = U f + K f
sin30º =
mgh =
1 2
mv
2
v = 2gh = 2(9.8m / s2 )(0.5m) = 3.1m / s
2.
For the problem above, instead of using energy methods, use Newton’s second law to find the acceleration of
the crate along the ramp and the equations of kinematics to determine the final speed of the cart.
 F = ma = mg sinq
a = gsin q = 9.8m / s2 sin30º = 4.9m / s2
v 2f = v2i + 2aDx
v f = 2aDx = 2(4.9m / s2 )(1m) = 3.1m / s
3.
A 1.50 kg water balloon is shot straight up with an initial speed of 3.00m/s. Neglecting resistive forces, what is
the maximum height of the balloon.
Ui + Ki = U f + K f
1 2
mv = mgh
2
v 2 (3.00m / s)2
h=
=
2 = 0.459m
2g 2(9.8m / s )
4.
When their car stalls on a level road, a group of students leap from the car and join in pushing it forward. The
cars mass is 1000 kg, and when they push it 5 m the car gains a speed of 2.0 m/s. What is its final kinetic
energy? How much work do they do in pushing it? If friction is negligible, what average force do they exert on
the car?
1 2 1
2
mv = (100kg )(2.0m / s) = 2000J
2
2
W = DK = K f - Ki = 2000J - 0 = 2000J
Kf =
W = FDx = 2000J
W 2000J
F=
=
= 400N
Dx
5m
5.
A 20.0 kg cannon ball is fired from a cannon at a muzzle speed of 1000.0 m/s and at an angle of 37.0º with the
horizontal. Find the maximum height reached by the ball.
Ui + Ki = U f + K f
1
1
2
2
mvi = mgh + mv f (still has x velocity at top)
2
2
v f = vi cosq
1
1
2
2
mvi = mgh + m(v i cosq )
2
2
1 2
1
2
vi = gh + (vi cos q )
2
2
2
(1000m / s)2
v
2
4
h = i (1- cos 2 q ) =
2 (1 - cos 37º ) = 1.85x10 m
2g
2(9.8m / s )
6.
A child of mass m takes a ride on an irregularly curved slide of height 6.00 m. The child starts from rest at the
top. Determine the speed of the child at the bottom, assuming no friction is present.
Ui + Ki = U f + K f
mgh =
1 2
mv
2
2
v = 2gh = 2(9.8m / s )(6m) = 10.8m / s
7.
For the child on the slide in the previous problem, assume a frictional force acts on the child. If the child’s final
velocity is 8.00 m/s and her mass is 20.0 kg, how much energy is lost to friction? Where does this energy go?
W = DK = Kno
friction
- K with
=
friction
1 2
1 2
mvno
- mvwith
2 friction 2 friction
1
1
2
m(vno2
- vwith
) = (20.0kg) (10.8m / s)2 - (8m / s)2 = 536J
2
2
friction
friction
Friction work is transformed into heat.
Note, you will get 526J, if you rounded 10.8 from previous problem.
W=
8.
[
]
A ball off mass m is dropped from a height h above the ground. Neglecting air resistance, determine the speed
of the ball when it is at a height y above the ground.
Ui + Ki = U f + K f
mgh = mgy +
1 2
mv
2
v = 2g(h - y )
9.
Determine the speed of the ball (in the previous problem) at y if it is given an initial speed vi at the initial height
h.
Ui + Ki = U f + K f
1
1
mgh + mvi 2 = mgy + mv f 2
2
2
1
1
gh + v i2 = gy + v f 2
2
2
1 2
2
g(h - y) = (v f - vi )
2
v f = 2g(h - y) + vi2
10. A pendulum consists of a sphere of mass m attached to a light cord of length L. The sphere is released from rest
when the cord makes an angle Q0 with the vertical, and the pivot at P is frictionless. Find the speed of the
sphere when it is at its lowest point.
Ui + Ki = U f + K f
q
L
1
mgh = mv 2
2
1
gh = v 2
2
L
Lcosq
1 2
v
2
v = 2gL(1 - cosq )
gL(1 - cosq ) =
11. A skier starts from rest at the top a frictionless incline 20.0 m long at an angle of 20.0º At the bottom of the
incline, the skier encounters a horizontal surface where the coefficient of kinetic friction between the skis and
snow is 0.210. How far does the skier travel on the horizontal surface before coming to rest.
Part1:
Ui + Ki = U f + K f
mgh = c
Part 2
K f for 1st part is K i for 2nd part.
K f for 2nd part is zero.
W = DK = K f - K i = -mgh
f k = mk N = mk mg
Wf = -mgh = fk Dx
- mgh = -m k mgDx
h = m k Dx
Dx =
h
20msin 20º
=
= 32.6m
Dx
0.210
12. The launching mechanism of a toy gun consists of a spring of unknown spring constant. When the spring is
compressed 0.120 m, the gun is able to launch a 35.0 g projectile to a maximum height of 20.0 m when fired
vertically from rest. Neglecting all resistive forces, determine the spring constant.
Ui + Ki = U f + K f
1 2
kx = mgh
2
2mgh 2(0.035kg)(9.8m / s2 )(20.12m)
k=
=
= 959N / m
(.12m) 2
x2
13. For the gun above, determine the speed of the projectile as it leaves the gun (i.e., when the spring is in its
equilibrium position at x=0.
Ui + Ki = U f + K f
1 2
1
kx = mgh + mv2
2
2
v=
kx2 - 2mgh
=
m
(959N / m)(.12m)2 - 2(.035kg)(9.8m / s2 )(.12m)
(.035kg )
= 19.8m / s
14. A mass of 0.80 kg is given an initial velocity vi =1.2 m/s to the right and collides with a light spring of force
constant k=50 N/m. If the surface is frictionless, calculate the initial maximum compression of the spring after
the collision.
Ui + Ki = U f + K f
1 2 1 2
mv = kx
2
2
x=
mv2
=
k
(0.8kg )(1.2m / s)2
50N / m
= 0.152m
15. A rocket is launched at an angle of 53º to the horizontal from an altitude h with a speed vi. Use energy methods
to find its speed when its altitude is h/2.
Ui + Ki = U f + K f
1
h
1
mgh + mvi 2 = mgÊ ˆ + mvi 2
Ë 2¯ 2
2
1
h
1
gh + v i2 = gÊË ˆ¯ + v i2
2
2
2
v f = vi2 + gh
16. If a 70.0 kg baseball player steals home by sliding into the plate with an initial speed of 10 m/s, how much
kinetic energy is dissipated by the frictional force stopping him?
1
DK = K f - Ki = - (70kg)(10m / s)2 = -3500 J
2
17. A 60.0 kg woman jumping from a window lands in an elevated fire rescue net 11.0 m below the window. She
momentarily stops when she has stretched the net by 1.50 m. Assuming that mechanical energy is conserved
during this process and that the net functions like an ideal spring, find the spring constant for the net.
Ui + Ki = U f + K f
1 2
kx
2
2mg(hi - h f ) 2(60kg )(9.8m / s2 )(12.5m)
k=
=
= 6533N / m
x2
(1.5)2
mghi = mgh f +
18. To pull a 50 kg crate across a frictionless floor, a worker applies a force of 210 N, directed 20º above the
horizontal. As the cart moves 3.0 m, what work is done on the crate by the worker’s force, the weight of the
crate, and the normal force exerted on the crate by the floor.
W = Fx Dx = (210N cos20º )(3m) = 592J
W = 0, perpendicular to motion
W = 0, perpendicular to motion
19. An initially stationary proton is accelerated in a cyclotron to a final speed of 3.0 x 106 m/s. How much work is
done on the proton? The mass of a proton is 1.67x10-27.
W = DK = K f - Ki =
1 2 1
2
mv f = (1.67x10 -27 kg)(3.0x10 6 m / s) = 7.5x10-15 J
2
2
20. A 2.0 kg block is dropped from a height of 40 cm on to a spring of spring constant k=1960 N/m. Find the
maximum distance the spring is compressed.
Ui + Ki = U f + K f
1 2
kx
2
1
mgh + mgx - kx 2 = 0
2
1
mgh + mgx - kx 2 = 0
2
1 2
kx - mgx - mgh = 0
2
1
(1960N / m)x 2 - (2kg)(9.8m / s2 )x - (2kg)(9.8m / s2 )(0.4m) = 0
2
mg(h + x ) =
980x 2 - 19.6x - 7.84 = 0
x = 0.1m