Sect. 7-4: Kinetic Energy; Work-Energy Principle. Energy : Traditionally defined as the ability to do work. We now know that not all forces are able to do work; however, we are dealing in these chapters with mechanical energy, which does follow this definition.
“Kinetic” Greek word for motion
An object in motion has the ability to do work.
Newton’s 2nd Law: Fnet= ma (1)
1d motion, constant a (v2)2 = (v1)2 + 2ad
a = [(v2)2 - (v1)2]/(2d) (2)
Work done: Wnet = Fnet d (3)
Combine (1), (2), (3):
Combine (1), (2), (3):
Wnet = mad = md [(v2)2 - (v1)2]/(2d)
Wnet = (½)m(v2)2 – (½)m(v1)2
DEFINITION:Kinetic Energy (KE)
(for translational motion; Kinetic = “motion”)
(units are Joules, J)
Wnet = K( = “change in”)
We’ve shown this for a 1d constant force. However, it is valid in general!
Wnet = K
The Work-Energy Principle
Note!:Wnet = work done by the net (total) force.
Wnetis a scalar & can be positive or negative (because K can be both + & -).If the net work is positive, the kinetic energy increases. If the net work is negative, the kinetic energy decreases.
The SI Units are Joulesfor both work & kinetic energy.
Wnet = K
This isNewton’s 2nd Lawin
Work & Energy Language!
For the hammer:
Wh = Kh = -Fd
= 0 – (½)mh(vh)2
For the nail:
Wn = Kn = Fd
= (½)mn(vn)2 - 0
A baseball, mass m = 145 g(0.145 kg) is thrown so that it acquires a speed v = 25 m/s.
a. What is its kinetic energy?
b. What was the net work done on the ball to make it reach this speed, if it started from rest?
Calculate the net work required to accelerate a car, mass m = 1000-kg, from v1 = 20 m/s to v2 = 30 m/s.
A car traveling at speedv1 = 60 km/hcan brake to a stop within a distanced = 20 m. If the car is going twice as fast,120 km/h, what is its stopping distance? Assume that the maximum braking force is approximately independent of speed.
Wnet = K = (½)m(v2)2 – (½)m(v1)2(Work-Energy Principle)
but, (v2)2 = 0 (the car has stopped) so-Fd = K= 0 - (½)m(v1)2
or d (v1)2
So the stopping distance is proportional to the square of the initial speed! If the initial speed is doubled, the stopping distance quadruples!
Note:K (½)mv2 0Must be positive, since m & v2 are always positive (real v).
A horizontal spring has spring constant k = 360 N/m. Ignore friction.
a. Calculate the work required to compress it from its relaxed length (x = 0) to x = 11.0cm.
b. A1.85-kg block is put against the spring. The spring is released. Calculate the block’s speed as it separates from the spring at x = 0.
c. Repeat part b. but assume that the block is moving on a table & that some kind of constant drag force FD = 7.0 N(such as friction) is acting to slow it down.
A block of mass m = 6 kg, is pulled
from rest (v0 = 0) to the right by a
constant horizontal force F = 12 N.
After it has been pulled for Δx = 3 m,
find it’s final speed v.
Work-Kinetic Energy Theorem
Wnet = K (½)[m(v)2 - m(v0)2] (1)
If F = 12 N is the only horizontal force,
then Wnet = FΔx (2)
Combine (1) & (2):
FΔx = (½)[m(v)2 - 0]
Solve for v: (v)2 = [2Δx/m]
(v) = [2Δx/m]½ = 3.5 m/s
A man wants to load a refrigerator onto a truck bed
using a ramp of length L, as in the figure. He claims
that less work would be required if the length L were
increased. Is he correct?
onto a truck bed using a ramp of
length L. He claims that less work
would be required if the length L
were increased. Is he correct?
NO!For simplicity, assume that it is wheeled up the on a dolly atconstant speed. So the kinetic energy change from the ground to the truck isK = 0. The total work done on the refrigerator is Wnet = Wman + Wgravity + Wnormal
The normal forceFNon the refrigerator from the ramp is at90º to the horizontal displacement & does no work on the refrigerator (Wnormal = 0).Since K = 0, by the work energy principle the total work done on the refrigerator is Wnet = 0. = Wman + Wgravity. So, the work done by the man isWman = - Wgravity.The work done by gravity isWgravity = - mgh[angle between mg & h is 180º & cos(180º) = -1]. So, Wman = mgh
No matter what he does he still must do the
SAME amount of work
(assuming height h = constant!)