Applied Physics I Newtons Laws Case Study

Published: 2021-07-13 03:25:05
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Newton’s First Law is the Law of Inertia. F = m * a F = force, m = mass, a = acceleration
An object will stay at rest (not moving) until some force is applied. An object in motion will stay in motion until some force is applied.
Games that have to do with balls or any kind of spheres are one example. If you are playing marbles the marbles do not move until they are hit by one of the other marbles or if a person pushes one of them or picks it up.
Newton’s Second Law of Motion Force on an object causes the object to move (in the direction of the force) at a certain acceleration which is a measure of distance and time. The amount of force used will be proportional to the amount of acceleration. When mass changes then acceleration changes too; a lighter mass allows an object to accelerate faster than a heavier mass which will slow the object down.
F = where F = Force, t = time and d = distance
An example of how this works is daily life is carrying things in my backpack. When I have a lot of books in my backpack they are heavy and I cannot walk very fast. When I have only a couple3 of notebooks in my backpack I walk faster because I am carrying less mass on my back.
Newton’s Third Law of Motion The Law of Action and Reaction
When two objects interact there is an action force which is the force exerted by the first object on the second object; the opposite force is called the reaction force and is the force the second object adds to the first object. ‘For every action there is an equal and opposite reaction.’
Playing baseball is an example of this law. In baseball a pitcher pitches the ball to a batter, when the batter hits the ball the ball will travel in the opposite direction equal to the magnitude the batter hit the ball with the bat.
Swimming laps in a pool is another example. A swimmer pushes against the wall with their feet when they make a turn at the end of the pool and then their body moves in an opposite direction in an equal reaction.
Variables:
mass = m = 10 kg
initial velocity = v1 = 5 m/s
final velocity = v2 = 4.8 m/s
time = t= 2 s
Force = ?N
Relationship:
deceleration a = and F = m * a
Computations:
a = = 0.2/2 = 0.1 m/s2
F = (10 kg) (0.1 m/s2) = 1 N
Variables:
Mass = m = 275 kg
Force = F = m*g
gravity = g = ? m/s
Relationship F = m*g
acceleration = 500 g
Force N = ?
Relationship: F = m* a
Computations:
a = 500 g = (500 g)(9.8 m/s2) = 4.9 N
 
Variables:
Forces A (0.5) B (3, -2) C (-1, 8) D (-7, -3)
Mass = m = 10 kg
Direction:
Acceleration = a
Relationships:
FN =
FNx = and FNy = ∑ Fy
tan φ =
FN = m * a
Computations:
FNx = = (10 + 3 – 1 – 7) = 5 N and FNy = = (5-2 + 8 – 3) = 8 N
FN = = 9.43 N
A = F/m + 9.43 N/0.10 kg = 0.943 m/s2
tan φ = = 8/5 = 1.6, therefore φ = arctan (1.6) = 57.99°

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