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CHAPTER 2.3: NEWTON’S 2ND LAW (F=M*A)
NAME _____________________________________ PERIOD ______ DATE ______________
Directions: Complete the following problems. Show work in the “Set-Up, Substitute, Solve (SSS) format.
1. A force of 20 N acts upon a 5 kg block. Calculate the acceleration of the object.
ans0401 (35.0K)
1. Newton's Second Law tells us how to calculate the acceleration if we know the force and the mass.
F=ma
We are interested in determining the acceleration, so we divide both sides of the equation by m to get
F/m=a
a = (20 N) / (5 kg)
2
2
a = 4 N / kg = 4 (kg m /s ) / kg = 4 m / s
2. An object of mass 300 kg is observed to accelerate at the rate of 4 m/s2. Calculate the force required to
produce this acceleration.
ans0402 (34.0K)
2. This problem also requires application of Newton's Second Law.
F=ma
2
F = (300 kg) (4 m / s )
2
F = 1200 kg m /s = 1200 N
2
since 1 N = 1 kg m / s .
3. A 5 kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 8 N
opposing the motion. Calculate the acceleration of the object.
ans0403 (39.0K)
3. The net force in the horizontal direction is the difference between the horizontal pulling force, F H, and the
frictional force, Ff, (we take the difference because the two forces act in opposite directions).
F = FH - Ff
F = 40 N - 8 N = 32 N
Now that we have the net force, Newton's Second Law is used to calculate the acceleration.
We use the same approach as was used for problem 1.
F=ma
a=F/m
a = 32 N / (5 kg)
2
a = 6.4 m / s
From the initial statement of the problem we know that the upward force of the table is equal to the weight of the
2
block, because there was no motion in the vertical direction, so the acceleration is 6.4 m / s in the direction of
the horizontal pulling force.
4. An object of mass 30 kg is in free fall in a vacuum where there is no air resistance. Determine the
acceleration of the object.
ans0404 (31.0K)
4. The only force on an object in free fall is the force of gravity (the object's weight). The acceleration of the
2
object is that due to gravity or a = g = 9.8 m / s .
5. An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50 newtons.
a. Determine the net force acting on the object and
b. calculate the acceleration of the object.
ans0405 (39.0K)
5. The weight of the object is found from the product of mass times the acceleration due to gravity
W=mg
2
W = (30 kg) (9.8 m / s ) = 294 N
The net force is the difference between the weight (directed downward) and the force due to air resistance
(directed upward).
F = W - FR
F = 294 N - 50 N
F = 244 N
The acceleration is calculated from the net force using Newton's Second Law as in previous problems as
a=F/m
2
a = (244 N) / (30 kg) = 8.13 m / s
6. A student pushes on a crate with a force of 100 N directed to the right. What force does the crate exert
on the student?
ans0406 (29.0K)
6. By Newton's Third Law, the crate pushes to the left on the student with a force of 100 N.
7. A force of 200 N is exerted on an object of mass 40 kg that is located on a sheet of perfectly smooth
ice.
a. Calculate the acceleration of the object.
b. If a second object identical to the first object is placed on top of the first object, what acceleration
would the 200 N force produce?
ans0407 (37.0K)
7. a) The acceleration is calculated using Newton's Second Law as
a=F/m
a = 200 N / 40 kg
2
a=5m/s
b) The force remains the same as in the first part of the problem but the mass is doubled so we obtain
a=F/m
a = 200 N / 80 kg
2
a = 2.5 m / s
Note that the acceleration on twice the mass by a given force is one half the acceleration calculated in the
first part of the problem.
8. Just before opening her parachute a skydiver of mass 50 kg reaches terminal velocity. Calculate the
force of air resistance.
ans0408 (36.0K)
8. The skydiver has reached terminal velocity. This means she is falling at constant velocity. From Newton's
First Law, if the velocity is constant the net force is zero. This means the force due to air resistance must be
equal to the weight of the skydiver. The weight of the skydiver is
W=mg
2
W = (50 kg) (9.8 m/s ) = 490 N (directed downward)
The force due to air resistance is therefore
FR = 490 N (directed upward)
9. For a person who has a mass 60 kg, calculate the weight in newtons and in pounds.
ans0409 (37.0K)
9. We know that weight is given by
W=mg
2
W = (60 kg) (9.8 m /s )
W = 588 N
From the list of Conversion Factors on the inside cover of the text we obtain the conversion that tells us that 1 lb
= 4.448 N, so we can convert 588 N to pounds by multiplying the right hand side of the equation by unity
expressed in the form
1 = 1 lb / 4.448 N
W = (588 N) (1 lb / 4.448 N)
W = 132.2 lb
10. An object of mass 10 kg is accelerated upward at 2 m/s2. What force is required?
ans0410 (42.0K)
10. The weight of the object is W = m g
2
W = (10 kg) (9.8 m / s ) = 98 N
On the free body diagram the weight is directed downward, and the external applied force, F e, is directed
upward. We can calculate the net force using Newton's Second Law as
F=ma
2
F = (10 kg) (2 m / s ) = 20 N
The external force and the weight act in opposite directions, so the net force is the difference between them or in
equation form,
F = Fe - W
We know the weight and the net force, so we can solve for the external applied force as
Fe = F + W
Fe = 20 N + 98 N
Fe = 118 N
Note that a force greater than the object's weight is required to accelerate it upwards. Part of the force is
required to overcome the force of gravity and part of it is required to give the desired acceleration. Compare this
problem to problem 2 where the motion is in a horizontal direction and the force of gravity was perpendicular to
the motion.