Definitions and Formulas
Acceleration of gravity
9.8m/s2
I struggle to move my desk. It doesn't move. How much work did I do?
No work is done because no distance was generated
5 things all levers must have
1. Effort
2. Load
3. Fulcrum
4. Load arm
5. Effort arm
I did 40J of work, but had to use 50J of energy. What is my efficiency?
Efficiency = work done divided by energy used x100%
40/50 x100 = 80%
Formula for force
F = ma
where m is mass (kg) and a is acceleration.
I move my desk 5m with a force of 25N. What is the work done?
125J
Explain how to remember what class a lever is.
FLE
123
The letter tells us what is in the middle, and the numbers tell us what class lever it is.
I did 250J of work, but it took me 300J of energy. What is my efficiency?
250/300 x 100% = 83.3%
Units of work
Joules
A battery has 450J of energy. The car it powers needs to travel 25m using 17.5N. Does the battery have enough energy to do the work?
Yes
25m x 17.5N = 437.5J, which is less than the 450J available
Torque of a 50N force on a 2m arm?
T = Fr
where F is force and r is the length of the arm.
T = 50Nx2m
= 100Nm
I moved a desk 2m using 50N of force. I had to use 240J of energy to do so. What is my efficiency?
Work done = Fxd = 50x2 = 100J
Energy used = 240J
Efficiency = 100/240 x 100% = 41.7%
How many times easier a simple machine makes our job, or load force divided by effort force
W =Fd
F = ma = 50x9.8N = 490N
d = 34m
W = 490N x 34m = 16660J
Find the torque of a 50kg person on a 1.5m arm.
T = Fr
F = 50kg x 9.8 = 490N
r = 1.5m
T = 490N x 1.5m = 735Nm
A toy car's battery has a limit of 500J. It was able to travel 120m with 3N of force before running out of energy. How efficient is the toy car?
Work done = Fxd = 3N x 120m = 360J
Energy used = 500J
Efficiency = 360/500 x 100% = 72%
Formula for efficiency
(work obtained / energy put in) x 100%
I did 5000J of work. If I used 25N of force, use your algebra skills and solve for distance.
W = Fd
5000J = 25d
5000J/25N = d
d = 200m
Two rhinos are on a seesaw.
First rhino: 625kg, 3m away from fulcrum
Second rhino: 3500kg, 0.55m away from fulcrum
Who is suspended in the air?
1. Calculate torque of each rhino
Rhino 1 = Fxr = (625x9.8)x3 = 18375 Nm
Rhino 2 = Fxr = (3500x9.8)x0.55 = 18865 Nm
2. Bigger torque means they will "win", and will push the seesaw down. Rhino 1 is suspended.
Two students use a 125J battery to power their science project robot.
The first student's robot moves an 8kg box up 1.2m before it runs out of energy.
The second student's robot moves a 7kg box up 1.35m before it runs out.
Which robot was more efficient?
First robot work = Fxd = (8kgx9.8)x1.2m =94.08J
First efficiency = 94.08/125x100% = 75%
Second robot work = (7kgx9.8)x1.35m = 92.61J
Second efficiency = 92.61/125x100% = 74%
First robot is more efficient.