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