Concepts A
Give a “field force” (a force that can act from range).
Gravity, electrostatic, or magnetic force
Give at least three contact forces.
Normal, friction, applied, tension
A 2 kg book rests on a flat surface. Solve for its
weight.
Fg = mg
Fg = (2kg)(9.81m/s2)
19.62 N
A 12 kg box is pushed at a constant speed of 4 m/s for 2 seconds. Solve for the net force acting on it while is was being pushed.
0 Newtons. The numerical information was irrelevant.
Constant speed = no acceleration, therefore
Fnet = ma = 0
A bat strikes a ball with a force of 100 N. How much force does the ball apply to the bat during this process?
100 N; 3rd Law
Describe the relative direction of a normal force to a surface.
Normal = Perpendicular
Identify all forces in this diagram.
F1 = Gravity
F2 = Normal
F3 = Friction
A 2 kg book rests on a flat surface. Solve for its
Normal force.
Fg = FN = mg
FN = (2kg)(9.81m/s2)
19.62 N
DAILY DOUBLE!
A 20 kg crate is lifted vertically upwards by a crane with a force of 500 N. Solve for the acceleration of the crate.
Fnet = ma = Fa - Fg
ma = Fa - mg
(20kg)a = 500N - 20kg(9.81m/s2
(20kg)a = 500N - 196.2 N
(20kg)a = 303.8 N
a = 15.19 m/s2
Which animal has the greater inertia, a 1,000 kg walrus moving at 2 m/s or a 70 kg cheetah moving at 30 m/s?
Inertia is dependent on mass, only.
Describe why (usually) more force is required to move an object than to keep it moving
Static friction is typically stronger than kinetic friction
An object weighs 10 N. What is its force of gravity from the Earth?
10 N
Weight is another term for the force of gravity from Earth.
A 5 kg box is resting on a 30o incline. Solve for its normal force.
FN = Fgy = mgcos(𝜃)
FN = (5kg)(9.81m/s2)(cos(30))
FN = 42.48 N
A 4 kg ball is hit with a force of 200 N from rest by a bat. The bat was in contact with the ball for 0.2 seconds. What is the ball’s velocity as it loses contact with the bat?
Fnet = ma
ma = 200 N
(4kg)a = 200 N
a = 50 m/s2
vf = vi + at
vf = 0 + (50m/s2)(0.2sec)
Vf = 10 m/s
An anxious student sits in a chair, worrying about an upcoming physics test. Identify the reaction force to their weight force.
The reaction to a person’s weight (also known as the force of gravity from the Earth) is the force of gravity they exert on the Earth. (Magnet on wall example)
DAILY DOUBLE!
Why does it typically require more force to lift an object than to push it across a level surface?
Gravity force is typically greater than friction force (𝜇 < 1)
True/False: The friction force acting on a moving object on a flat surface is affected by its mass. (Must have proper justification.)
True. Ff = 𝜇FN.
FN is dependent on mass, so Ff is as well.
A 5 kg box is resting on a 30o incline. Solve for the force of friction acting on it.
In this case, Ff = Fgx
Fgx = (5kg)(9.81m/s2)(sin30)
Fgx = 24.53 N
A 30 kg wooden crate is sliding down a 30o wooden incline (coefficient of friction = 0.3) . Solve for the force of kinetic friction acting on the crate.
Ff = 𝜇(mgcos(𝜃)
Ff = 0.3(30kg)(9.81m/s2)cos(30)
Ff = 76.46 N
A 0.01 kg mosquito is splattered on the windshield of a 4,000 kg truck on the highway. Compare the magnitudes of force on the mosquito to the force on the truck during this interaction.
Action forces and reaction forces are equal in magnitude; 3rd Law
A book rests on a level table. Describe what happens to the normal force acting on the book when a force is applied downward on it and explain why.
Normal force would increase by an amount equal to this applied force so that the book remains in force equilibrium.
List all factors that affect the coefficient of friction (𝜇).
The coefficient of friction is only dependent on what the materials in contact are made of.
A crate is resting on a 20o incline. Solve for the coefficient of its static friction force.
If at rest on incline; Fgx = Ff
mgsin(𝜃) = 𝜇(mgcos(𝜃)
𝜇 = tan(𝜃)
𝜇 = tan(20) = 0.364
A wooden crate is sliding down a 30o wooden incline. Solve its acceleration down the incline. (𝜇k = 0.3; mass is intentionally not given.)
Fnet = ma = Fgx - Ff
ma = (mgsin(𝜃) - 𝜇(mgcos(𝜃)
Use reference tables for 𝜇; mass cancels.
a =(gsin(𝜃) - 𝜇(gcos(𝜃)
a =(9.81sin(30) - 0.3(gcos(30)
a = 2.356 m/s2
A 3 kg rock and a 1 kg baseball are dropped. Each fall with the same acceleration (9.81m/s2). Use Newton’s 2nd Law to explain how this is possible.
Both objects have the same acceleration, but different masses. This means the Earth applied a different amount of force on each of the objects; the 3kg rock had 3x the force applied compared to the 1kg ball.