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Phys 101
CHAPTER 3 - EXPLAINING MOTION
Newton's Second Law
Newton's second law can be used to find the net force required to accelerate a given mass with a
given acceleration. The direction of the net force is the same as the direction of the acceleration.
The relationship can also be rearranged algebraically to solve for the acceleration or the mass.
Example: What net force is required to accelerate a 200-kg box with an acceleration of 4 m/s2
south?
The free body diagram (FBD) shows the single force Fnet that causes the
acceleration a of the box with mass m. We apply Newton’s second law to
the box,
Fnet = ma = 200 kg × 4m/s 2 = 800 kg ⋅ m/s 2 = 800 N
Thus the force F must have a size of 800 N and act toward the south, the same direction as
the acceleration.
Practice: What net force must be acting on a cyclist (and cycle) with a combined mass of 60 kg if
the acceleration is 2 m/s2 to the north? (Ans: 120 N north)
Example: What acceleration would result from a net force of 200 N acting westward on a person
with a mass of 80 kg?
Begin by solving Newton’s second law (Fnet=ma) for the unknown acceleration. Thus
F
200 N
N
a = net =
= 2.5
80 kg
kg
m
2
Recall that a Newton is equivalent to kg⋅m/s . So
kg ⋅ m/s 2
a = 2.5
= 2.5m/s 2
kg
The direction of the acceleration is toward the west.
Practice: What would the acceleration be if the force were increased to 300 N? (Ans: 3.75 m/s2
west)
Example: A spacecraft of unknown mass undergoes an acceleration of 3.0 m/s2 when a net force of
6000 N is applied to it. What is the mass of the spacecraft?
We begin with Newton's second law. To solve for the unknown mass, divide both sides by
the acceleration a to obtain.
Fnet m a
=
a
a
or
6000 N
kg ⋅ m s 2
F
m = net =
=
2000
= 2000 kg
a
3.0 m s 2
m s2
Practice: What would the mass be if it required 9000 N to produce the same acceleration?
(Ans: 3000 kg)
Phys 101
Weight
Weight is the force of gravity acting on an object. It is numerically equal to mg, were g is about
10 m/s2 here on the surface of the earth.
Example: What is the weight of a football player with a mass of 150 kg?
W = mg = 150 kg×10m/s 2 = 1500 N (; 330 lb)
Practice: What is the mass of a ballerina with a weight of 460 N? (Ans: 46 kg)
Example: What is the weight of an 80-kg person standing on Mars where the acceleration due to
gravity is 3.7 m/s2?
W = mg = 80 kg × 3.7 m/s 2 =296 N
Compare this with a weight of 800 N for the same person standing here on the Earth.
Friction and Air Drag Forces
Sliding friction and air resistance are two examples of dissipative force that always oppose the
motion of an object. You will want to construct a free-body-diagram showing all of the forces on
the object of interest. Pick a certain direction as the "positive direction" and then calculate the net
force on the object. After that, just follow Isaac Newton's instructions by setting the net force equal
to the mass of the object times its acceleration.
Example: A firefighter of mass 80 kg slides down a vertical pole with an acceleration of
4 m/s2. What is the force of friction on the firefighter?
Begin by choosing the positive direction as down. That seems reasonable since the
fire fighter will slide down the pole and not up the pole. The two forces on the fire
fighter are his weight, W=mg, which acts downward, the force of sliding friction, f,
which acts upward ( ie the friction force opposes the downward motion, so it must be
pointing upward). So the net force is
Fnet = W + (–f ) = W–f
Notice the – sign. It gives us the information about the direction of the friction force.
Substituting this expression for Fnet into Newton's second law, we have
Fnet = W–f = ma
Solving for f, the frictional force we have
f = W − ma
We can find the weight W=mg = 80kg×10m/s2 = 800N and the numerical value of ma =
80kg×4m/s2 = 320N. So the frictional force must be
f = W − ma = 800 N − 320 N = 480 N
Practice: Suppose the fire department bought a new and very slippery pole so that the
frictional force was reduced to f =240N. What would the acceleration of the same fire fighter
be now? (Ans: 7 m/s2)
f
W