Work is the product of the force and the distance through which the object moves W = F x d W = F x...
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Transcript of Work is the product of the force and the distance through which the object moves W = F x d W = F x...
WORK
In order for work to be done, three things are necessary:•There must be an applied force.•The force must act through a certain distance•The force must have a component along the distance.
Read the following statements and determine whether or not they represent examples of work.
A teacher applies a force to a wall and becomes exhausted.
A book falls off a table and free falls to the ground.
NO, displacement doesn’t occur.
Yes, displacement in the direction of force.
A truck carries a box in it’s bed 100 m.
NO, This is not an example of work. The force is upward on the box but the displacement is along the ground.
Is work being done or not?
• Mowing the lawn• Weight-lifting• Carrying groceries• Moving furniture up a
flight of stairs• Pushing against a
locked door• Swinging a golf club
• YES• YES• NO• YES
• NO
• YES
W = F d
W = Nm
W = (Newtons )(meters)
The units of work are;
W = Joule (J)
(In English units; W = foot poundsW = ft lb)
Steve lifts a 50 Nbox from the floor of the closet to a shelf 1.6 m high.
How much work does Steve do?
Given:F = 50Nm = 1.6m
Formula: W = F x d
Work: W = 50N x 1.6m
Answer:80 Joules
Matt lifts a 80 kg barbell upward for 1 meter at a constant speed, how much work does he do?
Given:m = 80kgd = 1mg= 9.8m/s2
First: What force must Matt provide ?
Formulas: W = F x d
F = m x g
Answer:784 J
Work: F = 80kg x 9.8m/s2 = 784N
W = 784N x 1m = 784 J
Work is the product of the force and the distance through which the object moves.
W = F x dW = F x d
Remember . . .
The SI unit of work is the Joule.
POWER is the rate at which work in done.
P = P = W W oror
t t
P = P = Fd Fd t t
The SI unit for power is the Watt (W).
1 Watt = 1 Joule/second
Steve lifts a 50 N box from the floor of the closet to a shelf 1.6 m high in 2 seconds.
How much work does Steve do? How much power does he use?
Given:F = 50Nm = 1.6mt = 2sec
Formulas: W = F x d P = W / t
Work: W = 50N x 1.6m = 80 JoulesPower: P = 80J / 2sec = 40 Watts
Matt uses 50 W of power to lift a barbell in 1.5 sec.
Given:P = 50Wt = 1.5sec
How much work does Matt do on the barbell?
Formula: P = W / t
Answer: W = 75 Joules
W = P x t = 50W x 1.5sec