Work Concepts

Work = F||*d = F.d
F = ma.
Power = W/t = Fv

1. Work is a scalar quantity. It is describes what is accomplished by the action of force when it acts on an object as the object moves through a distance. The unit of work is the joule.

2. a) If work is done by a constant force, work can be written as W = F||d b) To determine work done by a varying force, plot F cos theta vs distance and take the area under the curve.

3.
Energy is the ability to do work, this is also measured in joules.

4.
KE = 1/2mv^2. v = average velocity

5.
The work energy principle is that the net work done on an object is equal to the change in its kinetic energy.

6.
GPE = mgy. EPE = 1/2kx^2

7.
Conservative forces is when only initial and final positions matter, disregard path. Nonconservative forces is when the path it travels does matter.

8.
E = KE + PE. KE1 + PE1 = KE2 + PE2.E2 = E1.

9. The total energy is neither increased nor decreased in any process. Energy can be transformed from one form to another and transferred from one body to another, but the total amount remains constant.

10. –F(fr)d = 1/2mv^2 – 1/2mv^2 + mgy – mgy or 1/2mv^2 + mgy = 1/2mv^2 +mgy + F(fr)d.

Problem 1
A band of merry ducks want to ship a package of chickens with a mass of 2000kg upwards in an elevator 70 meters to the experimentation floor level. Unfortunately, these chickens happen to have been exposed to radiation (incidentally) and are mutating quickly. The elevator the chickens are being shipped in has a mass of 500kg, and the counterweight attached to the elevator has a mass of 750 kg. The ducks are in a rush to safely transport the chickens in 4 minutes. The ducks must aid in raising the elevator, as the counterweight will not be able to lift the elevator in time. What is the average power output the ducks must perform?

The mass of the elevator + the chickens is 2000kg + 500kg. The counterweight which "balances" the mass of the elevator is 750kg. Thus, the ducks must lift the mass of the elevator and chicken, minus the mass of the counterweight.
(2000kg + 500kg - 750kg)
1750 kg.

The force required to lift the elevator = mg.
1750g.

Now, we can calculate the power output, knowing the time and the distance and the force required.

P = W/t. P = (1750g*70)/(4*60).
P = 510W.

Problem 2
Find the work done by a duck exerting 500N of force in the direction 5i+j in moving a caged chicken from (0,0) to (1,2).

Start off by finding the force vector. Since the direction is 5i +j, the angle theta is atan(1/5) = 11.31 degrees.
The force vector is then 500(cos 11.31i + sin 11.31j) => 490.29i + 98.058j.
The displacement vector is i+2j.. Now take the dot product of the force and displacement.
F.d = (490.29*1) + (98.058*2) = 686.406J.

Problem 3
A backpack full of books weighing 53.0 N rests on a table in a physics laboratory classroom. A spring with a force constant of 150 N/m is attached to the backpack and pulled horizontally. If the spring stretches by 0.0210 m before the backpack begins to slip, what is the coefficient of static friction between the between the backpack and table?

hooke's law.. F = -kx.
In this case, F_friction = mu*mg*sin(theta).
Since it begins to slip after the spring stretches 0.0210 m, we can write -kx + mu*mg*sin(theta) = 0
-150 N/m * 0.0210m = -mu * 53.0 N * sin(90)
5.943396 x 10^-2 = mu.

Problem 4
A force F = 2.83 x^3 is applied to a mass of 1.81 kg for a time interval of 1.64 seconds. During this time interval, the mass moves from a position of x = 2.74 m to a position of x = 4.74 m. The force is colinear with the motion. What is the amount of work done by the force on the mass?

work = force*distance.
f = ma
We want to find the distance, basically, and the amount of force exerted over that distance.
since acceleration is the derivative of velocity..
work = int F . ds
work = (2.83/4)x^4 . (4.74-2.74)
let x = 1.64 seconds
work = 10.236J