Download Newton`s 2nd Law with Angled Forces Class Exercises Post

Newton’s 2nd Law with Angled Forces Class Exercises
Physics
1. A 0.55 kg kitten is laying on the floor when it grabs onto the end of a 2 m long piece of yarn, which
you start tugging toward you. Because of your tug, the yarn pulls the kitten with 5 N and the coefficient
of friction of friction between the floor and kitten’s fur is 0.3.
a) How quickly does the kitten accelerate?
[a = 6.15 m/s/s]
b) If you keep pulling with the same force, how fast will the kitten be going when it gets to you?
[vf = 4.96 m/s]
2. A 600 kg drag racing car crosses the finish line going 150 m/s, then hits the brakes and deploys a
parachute so that air resistance can help slow the car down. It takes 4.3 seconds for the car to stop.
The coefficient of friction between the tires and ground is 1.3.
a) How quickly does the car accelerate while it is braking?
[a = –34.88 m/s/s]
b) How strong is the air resistance on the car’s parachute as the car slows down?
[Fair = 13286.2 N]
3. Michaela goes out
skateboarding with her dog
(that is to say, she
skateboard while her dog
pulls her) on a flat road.
She is standing still when
she steps on her skateboard
and her dog begins to pull
her with 50 N at 20° below
horizontal. Michaela’s
mass (including the
skateboard) is 55 kg. The
coefficient of friction
against the wheels turning
is 0.08.
a) What is the net force acting on Michaela as her dog pulls her?
When we draw our free body diagram, we include the standard, common forces – gravity, normal
force, friction. We can calculate the force of gravity.
Fg = m∙g = 55 kg ∙ 9.8 N/kg = 539 N, down
Then we put in the force exerted by Michaela’s dog. Since that force is at an angle, we need to take
components:
Fdog x = Fdog ∙ cosθ = 50 N ∙ cos(20°) = 47.0 N, right
Fdog x = Fdog ∙ sinθ = 50 N ∙ sin(20°) = 17.1 N, down
We know that someone riding a skateboard on flat ground doesn’t move up or down, so the vertical
forces must cancel.
Fup = Fdown
FN = Fg + Fdog y
FN = 539 N + 17.1 N = 556.1 N, up
Ff = μ∙FN = 0.08 ∙ 556.1 N = 44.5 N, left
Now that we know all of the forces, we find the net force. Because the vertical forces cancel, we can
ignore them, so the net force is just the sum of the horizontal forces. We make right the positive
direction.
Fnet = Fdog x + Ff = 47.0 N, right + 44.5 N, right = +47.0 N + (–44.5 N) = +2.5 N = 2.5 N, right
b) How quickly will Michaela accelerate as her dog pulls her?
Fnet +2.5 N
a = m = 55 kg = +0.0455 m/s/s
c) How fast will Michaela be going when she reaches the end of the block, 100 m away from where she
started?
vf2 = vi2 + 2∙a∙d
vf2 = (0 m/s)2 + 2∙(0.0455 m/s/s)∙(100 m)
vf2 = 9.1 m2s2
vf = 3.02 m/s
4. As it starts moving, a tow truck pulls on the 1000 kg car attached to it with a force of 6000 N at 60°
above horizontal to the left. The coefficient of kinetic friction
against the motion is 0.05.
a) How quickly will the car accelerate?
[a = +2.77 m/s/s]
b) How long will it take for the car to get up to 12 m/s if the tow truck keeps pulling it?
[t = 4.33 s]
5. A crow is trying to fly away with a 2 kg bag of food that starts out sitting still on the sidewalk. It
pulls on the bag with 15 N of force at a 55° angle above horizontal toward the right. The bag slides 3 m
along the ground while speeding up to 1.4 m/s.
a) How quickly is the bag accelerating?
vf2 = vi2 + 2∙a∙d
(1.4 m/s)2 = (0 m/s)2 + 2(a)(3m)
1.96 m2/s2 = (6 m) ∙ a
a = 0.327 m/s/s
b) How much normal force does the ground exert on the bag?
The normal force is one of the vertical forces. Since the bag doesn’t come off the ground, the
vertical forces. That means we have to fill in the rest of the vertical forces and then solve for the
normal force:
Fg = m∙g = (2 kg)∙(9.8 N/kg) = 19.6 N
Fcrow y = 15 N ∙ sin(55°) = 12.3 N
Fup = Fdown
FN + Fcrow y = Fg
FN + 12.3 N = 19.6 N
FN = 7.3 N
c) How much friction does the sidewalk exert on the bag?
There is an unbalanced (net) force on the bag, which is what makes the bag accelerate.
Fnet = m∙a = (2 kg)∙(0.327 m/s/s) = 0.654 N
The net force is created by all of the horizontal forces – the horizontal component of the crow’s force
and the force of friction. The force of friction has to be just right. We’ll use positive to symbolize
right.
Fnet = Fcrow x + Ff
0.654 N = 8.6 N + Ff
–7.946 N = Ff
d) What is the coefficient of friction between the ground and the bag?
Ff 7.946 N
μ =F = 7.3 N = 1.09
N
6. After an explosion in the hold of the Mary Celeste, the captain and crew climb off of the boat and into
the lifeboat for safety. They tie the 600 kg lifeboat to the back of the Mary Celeste with a rope which
pulls the lifeboat forward at a CONSTANT SPEED of 3 m/s against the water resistance which tries to
stop them. The rope pulls with a force of 230 N at 30° above horizontal
a) How strong is the water resistance against the lifeboat?
[Fwater = 199.2 N, backward]
b) How hard does the water have to lift to hold the boat up as it floats there?
[Fbuoyant = 5765 N]