Abeka Physics Section 11.1 Flashcards
Define and describe work
> work - “transfer of energy from one body to another”
if it fails to produce motion, it has failed to do work
absence of motion, no every is transmitted from your body to things acted upon: therefore, no work is done
force resulting in motion allows transfer of energy from one object to another and so accomplishes work
Describe the relation between work and force
> amount of work accomplished depends on magnitude of force
greater force, more energy transferred, and more work done
motion can be inhibited by opposing force–friction (rubbing of object against floor agitates molecules at points of contact and raises temperature of both surfaces. thus, part of applied force produces heat energy, while remaining part–net force in direction of motion–is spent on accelerating object and giving it kinetic energy. as applied force increases, total energy imparted to trunk and floor also increases; consequently, more work is accomplished
Describe how the basic work equation is derived
> when force causes motion, energy is transferred from one object to another, and work is done
when force applied over given distance, amount of work is proportional to strength of force
amount of energy transferred in equal intervals of distance is same if force is same
as constant force causes something to move over increasing distance, there is proportional increase in energy transfer and work
proportionalities 1 and 2 can be combined into single equation: W = Fd
work done on object is equal to product of force applied to object in direction of motion and distance object moves
Give the units of work
> SI unit of work: joule (J)
>US customary system unit of work: foot-pound (ft x lb)
Describe an explain how work is a scalar quantity
> although work is product of 2 vectors, force and distance, work is scalar quantity because direction in which it is accomplished is irrelevant
stages of work should not be treated as vectors
if work were vector, then total work of 2 efforts, equal in magnitude but opposite in direction, would incorrectly represent work’s magnitude
Solve this: A man pushes a 100.-kg crate 20.0m across horizontal floor at a constant speed. If coefficient of kinetic friction is 0.65, how much work is done on the crate?
Since crate is moving at constant speed, the applied force must be just sufficient to overcome friction. Thus, the applied force must be equal in magnitude to the friction: F = F(sub kf). Since
F(sub kf) = u(sub k) x F(sub n) and the crate is on a level floor, the normal force equals the magnitude of the weight. Since
F(sub w) = mg,
F = F(sub kf) = u(sub k)mg
Substitute into equation 3 to obtain
W = u(sub k)mgd
Therefore, since u(sub k) is 0.65, m is 100.0 kg, g is 9.80 m/s^2, and d is 20.0 m,
W = (0.65)(100.kg)(9.80m/s^2)(20.0m)
W = 13,000 J
Explain how to solve for W with force applied at an angle
> if force applied at angle to direction of motion, only component of force in direction of motion determine acceleration
at same time, only this same component (in direction of motion) determines amount of work done
if angle between force and motion is theta, work equation becomes W = (F cos theta)d or W = Fd cos theta
Solve this: A girl is pulling a sled that is accelerating at a rate of 5.00m/s^2. Neglect the friction between sled runners and snow. The mass of the sled is 5.00 kg, and the angle between the rope and the sled is 30.0 degrees. (a) How much tension force is in the rope? (b) How much work is the girl doing on the sled when she pulls it 2.00 m?
(a) Since only the force in the direction of motion determines acceleration, you need to find F(sub x), the horizontal component of the tension. Recall that F(sub x) = F cos theta
Substitute this into F = ma and use the horizontal acceleration to obtain
F cos theta = ma
Solving for F, F = ma/ cos theta
where m is 5.00 kg, a is 5.00 m/s^2, and theta is 30.0 degrees. Substitution gives
F = (5.00 kg)(5.00 m/s^2)/ cos 30.0 degrees
F = 28.8675… N = 28.9 N
Therefore, the rope has a tension of about 28.9 N.
(b) Work is found from equation 4. The tension in the rope is 28.8675… N, and d is 2.00m. Substitution gives
W = Fd cos theta
W = (28.8675…N)(2.00m)(cos 30.0 degrees)
W = 50.0 J
If a man carries a box at a constant speed along a horizontal path instead of pushing it, he does no work. Explain this.
> he did some work to start box moving, but once he is underway, work is practically zero (except for force required to overcome air resistance)
reason is that he applies all of his force under the box at right angles to motion
therefore, if the force he applies is F, the component of force in the direction of motion is F cos 90.0 degrees = F(0) = 0
Describe concurrent applied forces
> work is done by each separate force contributing to motion
if 3 people push a piano in same direction with individual forces of 50 N, 90 N, and 30 N
if they succeed in moving piano 15m, total work is
(50.0 N + 90.0 N + 30.0 N)(15.0 m) = 2550 J
the share done by each person is found by multiplying force he uses by how far piano moves
thus, 3 people separately do 750, J, 1350 J, and 450. J of work respectively, but notice that sum of individual work equals total work
Define and Describe negative work
> force applied opposite to motion does negative work
example: any sort of friction
another example: father and son working together to clean out attic. father lowers box of books through hole in ceiling to son, who is standing on ladder. son then climbs down ladder and deposits box on floor. he feels as if work is backbreaking, but all he has done is negative work. force he has exerted has been upward, in opposition to downward motion of box. positive work has been done by gravity
“negative” aspect of negative work indicated that energy is flowing in his direction: energy appears mostly as heat energy or potential energy in his muscles
Define and describe net work
> as weight is lifted at constant speed, it has no net force acting upon it; thus, upward force applied by weight lifter is equal to downward force of gravity
similarly, as weight is lowered at constant speed, upward force and downward force of gravity are same
upward force must exceed gravity to start weight upward, and gravity must exceed upward force to start weight downward; but over whole distance that weight moves up and down, force applied by weight lifter is almost constant
thus, positive work of lifting is only slightly more than negative work of lowering
combining work of lifting and work of lowering gives net work done by weight lifter
net work - algebraic sum of positive work and negative work
suprisingly, weight lifter’s net work is little more than zero
this is why less net work is done when you climb to third floor and return to second than when you climb to third floor and stay here
Describe work and potential energy
> when push box at constant speed, work is done
energy is transferred from your body to box and its surroundings specifically, it gains kinetic energy and produces heat energy at surfaces in contact with each other
when work is done to lift weight at constant speed, outcome is increase in neither kinetic energy nor heat energy
lifted weight does, however, acquire gravitational potential energy
this reaches maximum at top of weight’s path, where weight has farthest to fall
what is the special rule applied when lifted weight acquires gravitational potential energy
> special rule applies when, as an object is moved by force, energy spent on object is budgeted so that all is used to increase potential energy of object and none is wasted on overcoming friction
in such cases, amount of work accomplished does not depend on path taken, provided final distance from zero or reference position is same
neglecting friction, whether you mount platform by leaping onto it or by climbing steps or by walking up ramp, you do same amount of work
