Problems Involving Forces
Frictional Force
A smooth wooden block is placed on a smooth
wooden tabletop. You find that you must
exert a force of 14.0 N to keep the 40.0 N
block moving at a constant velocity.
a) What is the coefficient of sliding friction for
the block and the table?
b) If a 20.0 N brick is placed on the block, what
force will be required to keep the block and
brick moving at a constant velocity?
Frictional Force
FN
REMEMBER:
If the object is moving at a constant
velocity, then the force applied (FA)
will equal the Frictional force (Ff)
FA
W
Ff
Ff = µFN
Part a:
Given:
EQN:
So µ = Ff/FN
FA= Ff = 14.0 N
Ff = µFN
µ = 14.0 N/40.0 N
W = -40.0 N
FN = 40.0 N
µ=?
µ = .350
Frictional Force
REMEMBER:
If the object is moving at a constant velocity, then the
force applied (FA) will equal the Frictional force (Ff)
Part B:
New weight:
W = -40.0 N + (- 20.0 N) = -60.0 N
Therefore, the Normal Force:
FN = 60.0 N
Constant Velocity makes:
FA = Ff
Ff = µFN
so
FA = µFN
FA = (.350)(60.0 N)
FA = 21.0 N
Force and Acceleration
A box weighs 75 N.
a) What is the mass of the box?
b) What is the acceleration of the box if an
upward force of 90 N is applied?
Force and Acceleration
a) What is the mass of the box?
Given:
W = -75 N
g = -9.8 m/s2
W = mg → m = W/g
m = -75N/-9.8 m/s2
m = 7.7 kg
Force and Acceleration
b) What is the acceleration of the box if an
upward force of 90 N is applied?
Since the question is asking for acceleration, we need to
use Newton’s 2nd Law, Fnet = ma
Given:
W = -75 N
Upward force = 90 N
m = 7.7 kg
Fnet = ma → a = Fnet/m
a = (-75 N + 90 N)/7.7 kg
a = 15 N/7.7 kg
a = 1.9 m/s2