Download Newton`s 2nd Law on Hills Class Exercises Answers

Newton’s 2nd Law on Hills Class Exercises
Physics
1. Jessica gets a chance to ride down a luge course, so she quickly gets on the sled and pushes herself so that
she is going 5 m/s when she starts to slide down the 7° slope. Jessica and the sled mass 60 kg. The coefficient
of friction of the sled on the ice is 0.05. She makes it down the course in 51 s.
a) How quickly does she accelerate as
she rides down the luge course?
She is accelerating because the
forces acting on her are not
balanced. First, we need to figure
out how big all of the forces acting
on her are and then we can see how
they combine into the net force. I
start by drawing my free body
diagram, being sure to make the
normal force perpendicular to the
surface. Then I start calculating
my forces.
Fg = m∙g = 60 kg ∙ 9.8 N/kg = 588 N, down
Remember that on a hill, we change our reference frame so that we use parallel (uphill-downhill) and
perpendicular (into the hill-out of the hill) as our cardinal directions (instead of vertical and horizontal).
This has the advantage of making every force except gravity go in a cardinal direction. We are left with
taking components of gravity. I draw in my parallel and perpendicular components (as shown in the
diagram) and calculate their size.
Fg⊥ = Fg∙cosθ = 588 N ∙ cos(7°) = 583.6 N, into hill
Fg∥ = Fg∙sinθ = 588 N ∙ sin(7°) = 71.7 N, downhill
Because she is not going into the hill or flying off of it, I know that the forces in the perpendicular direction
have to be balanced:
FN = –Fg⊥ = 583.6 N, out of hill
Ff = μ∙FN = 0.05 ∙ 583.6 N = 29.2 N, uphill
Since the perpendicular forces cancel, only the parallel forces contribute to the net force. I will call
downhill the positive direction.
Fnet = Fg∥ + Ff = 71.7, downhill + 29.2 N, uphill = +71.7 N + (–29.2 N) = +42.5 N = 42.5 N, downhill
It is the net force that makes her accelerate:
Fnet 42.5 N
a = m = 60 kg = 0.708 m/s/s
b) How long is the luge course she rides?
d = vi∙t + ½∙a∙t2 = +5 m/s ∙ 51 s + ½ ∙ (+0.708 m/s/s) ∙ (51 s)2 = 1176 m
c) How fast is she going at the bottom of the luge course?
vf = vi∙t + a∙t = (+5 m/s) + (+0.708 m/s/s) ∙ 51 s = 41.1 m/s
2. 70 kg Ben sits at the top of a 150 m long, 30° angled water slide for a few seconds before letting go. The
coefficient of friction between Ben and the wet concrete is 0.1.
a) How quickly does Ben accelerate as he slides down the waterslide?
[a = 4.05 m/s/s]
b) How fast is the Ben going when he reaches the bottom of the slide?
[vf = 34.9 m/s]
3. A 0.03 kg pinball is hit by one of the paddles and comes off going up the 10° slope at 1.1 m/s. The
coefficient of friction against the pinball’s rolling is 0.04.
a) How quickly does the pinball accelerate as it rolls up the slope?
[a = –2.09 m/s/s]
b) How high up the slope does the pinball get before it stops and rolls back down?
[d = 0.29 m]
4. NASA designs a rail gun, which uses
magnets to pull a spaceship up a 50°
ramp and launch it into space. When the
rail gun is turned on, the 120,000 kg
spaceship is sitting at the bottom of the
2000 m long tube. When the spaceship
leaves the railgun it is going 780 m/s.
The coefficient of friction between the
spaceship and the launch tracks is 0.1.
a) How quickly does the spaceship
accelerate while it is being launched?
I pick uphill to be the positive
direction, so both distance and final
velocity are positive.
vf2 = vi2 + 2∙a∙d
(780 m/s)2 = (0 m/s)2 + 2∙a∙2000 m
608400 m2/s2 = 4000 m ∙ a
152.1 m/s/s = a
b) How strong is the force of the
magnets on the spaceship while it is
being launched?
The space ship is accelerating
because the forces on it are not
balanced – the amount of imbalance is called the net force and we can use Newton’s 2nd Law to find it:
Fnet = m∙a
Fnet = 120,000 kg ∙ (+152.1 m/s/s)
Fnet = +18,252,000 N
Fnet = 18,252,000 N, uphill
The net force acts to push the spaceship uphill – in the parallel direction. That tells us two things: 1) that the
perpendicular forces cancel out and 2) the net force is made up of the forces in the parallel direction.
Fnet = Ff + Fg∥ + Fmagnet
I can use that fact to solve for the force exerted by the magnet if I can find all of the other forces that
contribute to the net force.
Fg = m∙g = 120000 kg ∙ 9.8 N/kg = 1,176,000 N, down
Fg⊥ = Fg∙cosθ = 1,176,000 N ∙ cos(50°) = 755,918 N, into hill
Fg∥ = Fg∙sinθ = 1,176,000 N ∙ sin(50°) = 900868 N, downhill
FN = –Fg⊥ = 755,918 N, out of hill
Ff = μ∙FN = 0.1 ∙ 755,918 N = 75,592 N, downhill
Now we can put all of our information together to find the missing force. We pick uphill
Fnet = Ff + Fg∥ + Fmagnet
18,252,000 N, uphill = 90087 N, downhill + 900868 N, downhill + Fmagnet
+18,252,000 N = –75592 N + –900868 N + Fmagnet
+19,228,460 N = Fmagnet
so
Fmagnet = 19,228,460 N, uphill
5. A 400 kg roller coaster car gets to the top of the 60° big hill going 4 m/s. The car reaches the bottom of the
hill 3.4 s later going 27 m/s. The whole way down, the riders can feel the push of the air against them and the
car. The coefficient of friction is 0.15.
a) How quickly does the roller coaster car accelerate down the big hill?
[a = 6.76 m/s/s]
b) How strong is the force of air resistance against the car as it rolls down the hill?
[Fair = -397 N]
6. A team of oxen pull a 1500 kg covered wagon up a 15° hill at a constant speed of 2 m/s. The coefficient of
friction against the wagon’s wheels is 0.3.
a) How quickly does the wagon accelerate as it rolls up the hill?
[a = 0 m/s/s]
b) How large a force do the oxen exert on the wagon?
[Foxen = 8064.4 N]