CK-12-Physics - Intermediate

http://www.ck12.org Chapter 6. Work and Energy

FIGURE 6.12

Answer:Work is defined asW= (Fcosθ)x. Since the friction force opposes the motion of the tire it acts oppositely

to the tire’s displacement. The angle,θbetween the force of friction and the displacement is 180-degrees, and the

cos 180◦=−1.

Recall that the displacement of the tire was 3.25 m and the force which the man applied in the direction of the tire’s

motion was 272 N. The force of friction therefore must have been -272 N:fk=Fcos 180◦= 272 (− 1 ) =− 272 N.

The work done by friction is therefore(− 272 )( 3. 25 ) =− 884 J.

The work done by friction is transformed into heat energy. (The bottom of the tire is warmer than its surroundings.)

Eventually, the heat dissipates but it does not “disappear.” The work due to friction is transformed into thermal

energy, which increases the internal energy of the universe. Energy can neither be created nor destroyed, but it can

be transformed from one form of energy into another form of energy.

Potential Energy

In understanding force, we know that an object close to the surface of the Earth experiences a force equal tomg,

whereg, is called the intensity of the field. In introductory physics,gis more commonly known as the gravitational

acceleration and has an average value of approximately 10 m/s^2.

When an object is lifted in the gravitational field, work is done upon it since a forceFlifts the object a heighth. See

Figure6.13. IfF=Mg, the work the forceFdoes on the object in lifting it a heighthisMgh.

Lifting the object increases its potential energy byMgh. In general, we establish an arbitrary zero point for the

potential energy. This zero point is usually set at the lowest level at which the object finds itself in the situation

described.

InFigure6.13, suppose you lifted a volleyball from the ground to over your head. The volleyball has a weight

of mg=3N and the height it is lifted to ish= 2 .0m above the ground. Your lifting forceFhas done 6J of work

upon the object.

http://demonstrations.wolfram.com/PotentialEnergyOfObjectsFromDailyLife/

Check Your Understanding

1a. What is the net work done on the object inFigure6.13?
Answer:The net force,∑Fon the object is zero since gravity is exactly matched by the lifting force. We know this
because the ball is not accelerating - it is only lifting at a constant speed. Since the net force is zero, the net work,
W= (∑F)h=F h−mgh=0.

1b. True or False: Gravity does negative work on the object inFigure6.13.