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Newton's Second Law
Newton's 2nd Law Lab
The purpose of this lab was to investigate the relationship between the net force on an object and the acceleration of the object. To do this, we put a cart on a low friction ramp and hooked a force sensor(that was plugged into Labquest) to the cart. We pushed and pulled the cart 5 times and the Labquest created two graphs, an acceleration vs. time graph and a force vs. time graph. The two graphs looked very similar. This showed that the net force and the acceleration on an object are directly proportional. If one increases, the other will also increase (and vice versa). Since the two are directly proportional, they would produce a linear graph if they were graphed together.
M = Fnet/A
N = kg(m/s^2)
N/kg = m/s^2
A = Fnet/Mass
This is where we get our equation A=F/M.
Fnet = AM
a
(m/s^2)
* force doubles; acceleration doubles
* mass double; acceleration is halved
* mass and force doubles; acceleration stays the same
* If you have two opposite directions, one must be negative for the equation to be correct*
f (N)
A. What is the net force on the object?
Step 1: Add all forces together for forces x and y
Fn = 20N
Fx = 10N = Fnet
Fx = Ft = Fnet
What is the acceleration of a falling book if it has a mass of 10kg and produces a force of 4N?
F=4N
M=10kg
A=?
Fy = -20N + 20N = 0
Fy = Fg + Fn = 0
A=Fnet/M
A=4N/10kg
A= 0.4m/s^2
Ft = 10N
B. What is the acceleration of this object?
most acceleration; most net force
V
d
* to find mass from weight on Earth (Fg), divide the weight (Fg) by 10m/s^2 *
c
d
Fg = 20N
e
What is the mass of a falling rock if it produces a force of 147N?
F=-147N
M=?
A=-10m/s^2
Mass= 20N/10(m/s^2)
Fnet
a
b
___
A=
M = Fnet/A
M = (-147N)/(-10m/s^2)
M = 14.7kg
Mass = 2kg
c
b
M
T
What net force is required to accelerate a car at a rate of 2m/s^2 if the car has a mass of 3,000kg?
F=?
M=3,000kg
A=2m/s^2
Fnet = MA
Fnet = 3,000kg(2m/s^2)
Fnet = 6,000N
e
d
A= 5m/s^2
10N
c
b
___
A=
greatest change of speed (forces most unbalanced)
a
2kg
no speed change
(most balanced forces)