By: Ryan Carpenter
formulas
-E = fd
-PE = mgh
-force of friction F = uf
1) A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant force of 45 N. How far must the student be pushed, starting from rest, so that her final kinetic energy is 352J? TOP
E = fd
352 = 45d
352/45 = d
d = 7.822222222
d = 7.8 m
2 A 2E3 kg car accelerates from rest under the actions of two forces. One is a forward force of 1140 N provided by traction between the wheels and the road. The other is 950 N resistive force due to various frictional forces. Use the work-kinetic energy theorem to determine how far the car must travel for its speed to reach 2.0m/s? TOP
so before we have pure kinetic energy and after, all that energy goes into friction
to find the net force, find big – small = 1140 – 950 = 190 N
.5mv^2 = fd
.5(2E3)(2.0)^2 = (190)d
.5(2E3)(2.0)^2/190 = d
d = 21.052631158
d = 21 m
3) A 2.1E3 kg car starts from rest
at the top of a driveway that is sloped at the angle of 20.0 degrees with the
horizontal. An average friction force
of 4E3 N impedes the car’s motion so that the car’s speed at the bottom of the
driveway is 3.8 m/s. What is the length
of the driveway? TOP
before = potential
after = kinetic + friction
so to find potential, use trig,
sin () = opp/hyp
sin () = h/l
solve for the height in terms of length
h = l sin 20
now plug it in
mgh = fl + .5mv^2
mg (l sin20) = fl + .5mv^2
solve for l
mg lsin20 –fl = .5mv^2
l(mgsin20 – f) = .5mv^2
l = .5mv^2 / (mgsin20 – f)
l = 15162/3045.956973
l = 4.977745955
l = 5.0 meters
4) A 75 kg bobsled is pushed along a
horizontal surface by two athletes after the bobsled is pushed a distance of
4.5m starting from rest, its speed is 6.0 m/s.
Find the magnitude of the net force on the bobsled. ` TOP
before = fd
after = .5mv^2
fd = .5mv^2
f(4.5) = .5(75)(6.0)^2
f = .5(75)(6.0)^2/(4.5)
f = 300 N
5) A 10.0 kg crate is pulled up a rough
incline with an initial speed of 1.5 m/s.
The pulling force is 100.0 N parallel to the incline, which makes an
angle of 15.0 degrees with the horizontal.
Assuming the coefficient of kinetic friction is .40 and the crate is
pulled a distance of 7.5 m, find the following. TOP
a.
the work done by the Earth’s gravity on the crate
first we must find the height it goes up to
sin 15 = h/l
lsin15 = h
7.5sin15 = h
h = 1.941142838
now the potential energy
E = mgh
E = (10)(9.81)(1.941142838)
E = 190.426, pointing down so the force will be negative
E
= -190J
b. the work done by the force of friction on the crate
find the force perpendicular to the crate
downward force = 9.81*10
98.1/cos15 = perp
friction = u(perp force)
friction = .40 * 98.1/cos15
friction = 37.72N
force = fd
force of friction = 37.72*7.5
force of friction (impeding) = -282.94
F = -280J
c. the work done by the puller on the crate
E = fd
E = (100)(7.5)
E = 750J
d. the change in kinetic energy of the crate
do part e first
then initial kinetic energy – final
11.25 – (.5)(10)(7.6)^2 = change in kinetic
change = 277.55
change = 280J
e. the speed of the crate after it is pulled 7.5 m
before = kinetic
after = kinetic + potential + friction + puller
.5mv^2 = .5mv’^2 + mgh + fd + fd
plug in above answers/ do some simple calculations
11.25 = 5v^2 + -190 + -280 + 750
-268.75 = 5v^2
get rid of the negative to make it nice
(268.75/5)^.5
speed after = 7.6 m