Download Newton`s Second Law Problems

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Newton’s Second Law Problems
When an unbalanced force (or net force) is exerted on an object, the object accelerates in
the direction of the force. The acceleration is proportional to the net force and inversely
proportional to the mass of the object,
Net Force = Mass * Acceleration
Fnet = ma
1. Butch, the 72kg star quarterback, collides with Trask, a stationary left tackle and
is brought to a stop with an acceleration of –20m/s2.
A. What net force acts to stop Butch? Show all your work.
a = -20 m/s2
m = 72 kg
FNET = ?
FNET = m • a
FNET = (-20 m/s2) • (72 kg)
FNET = - 1440 N
B. Sketch a velocity-time graph (no numbers, just show the shape) for Butch’s
motion.
v
t
2. While playing with the Boston Red Sox, David Ortiz swung at a 0.15kg baseball
and accelerates it at a rate of 3000m/s2.
A. What was the net force (exerted by the bat) acting on the ball? Show all the
steps.
a = 3000 m/s2
m = 0.15 kg
FNET = ?
FNET = m • a
FNET = (3000 m/s2) • (0.15 kg)
FNET = 450 N
B. Draw a force diagram for the ball while it is being hit by the bat. Include the
net force.
FN
FNET = 450 N
FG
3. Claudia stubs her toe on the coffee table with a force of 100N. If Claudia’s foot
has a mass of 1.8kg, what acceleration does her foot experience?
a=?
m = 1.8 kg
FNET = 100 N
a = FNET / m
a = (100 N) / (1.8 kg)
a = - 55.6 m/s2
Net force: The sum of all forces acting on an object. Direction must be considered! Also,
we will generally look at the net force in the left-right and the net force in the up-down
direction separately.
1. A skydiver is falling toward earth. His weight (the force that the earth exerts on
him) is 1000N. Before he opens his parachute there is 400N of air resistance force
acting on him.
a. Draw a quantitative (that means with numbers) force diagram for the
skydiver. What is the net force acting on the skydiver?
FAR = 400 N
FNET = 600 N down
FG = 1000 N
b. If his mass is 100 kg, what is the skydiver’s acceleration? In what
direction?
a=?
m = 100 kg
FNET = 600 N
FAR
a = FNET / m
a = (600 N) / (100 kg)
a = 6 m/s2 down
2. The same skydiver opens his parachute. The force of air resistance is now 1200N.
a. Draw a new force diagram (yes, make it quantitative) for the skydiver.
= 1200 N
What is the net force acting on the skydiver now?
FG = 1000 N
FNET = 200 N up
b. If his mass is still 100 kg (did you think it might change?), what is the
skydiver’s acceleration? In what direction?
a=?
m = 100 kg
FNET = 200 N
a = FNET / m
a = (200 N) / (100 kg)
a = 2 m/s2 up
3. You see a skydiver falling toward Earth. You know that his weight is 800N and
his mass is 80kg. If you measure his acceleration to be 2 m/s2 downward, how
much air resistance is acting on him?
a = 2 m/s2 down
FNET = m • a
FAR = ?
2
m = 80 kg
FNET = (80 kg) • (2 m/s )
FNET = ?
FNET = 160 N down
FG = 800 N
Looking at the FBD on the side and the net force above:
FNET = FG – FAR
160 N = 800 N + FAR FAR = - 640 N or 640 N UP