# Exercise 5C| 1 | 2 | 3 | 4 | 5 | back to g solutions | physics home|

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