1. Problems Involving Forces

2. 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?

3. Frictional Force Ff = µFN So µ = Ff/FN µ = 14.0 N/40.0 N µ = .350
REMEMBER:
If the object is moving at a constant velocity, then the force applied (FA) will equal the Frictional force (Ff) FA Ff W
Ff = µFN
So µ = Ff/FN
µ = 14.0 N/40.0 N
µ = .350
Part a:
Given: EQN:
FA= Ff = 14.0 N Ff = µFN
W = N
FN = 40.0 N
µ = ?

4. Frictional Force REMEMBER: FA = (.350)(60.0 N) FA = 21.0 N
If the object is moving at a constant velocity, then the force applied (FA) will equal the Frictional force (Ff)
Part B:
Ff = µFN so FA = µFN
FA = (.350)(60.0 N)
FA = 21.0 N
New weight:
W = N + ( N) = N
Therefore, the Normal Force:
FN = 60.0 N
Constant Velocity makes:
FA = Ff

5. 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?

6. 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

7. 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