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Wize AP Physics C: Mechanics Textbook > Unit 3: Work, Energy, and Power (14-17%)

Work-Energy Theorem

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Work-Energy Theorem


An object that moves with velocity v has kinetic energy calculated by:

K=12mv2\boxed{K=\frac{1}{2}mv^2}

  • Kinetic energy is a scalar quantity and measured in Joule.
  • Kinetic energy is also shown by KE or  EkKE \ \text{or}\ \ E_k.

Work-Energy Theorem

The work done by all forces acting on a body is equal to the change in the body's kinetic energy.

Wnet=ΔK = 12mvf2 − 12mvi2\boxed{W_{net}=\Delta K\ =\ \frac{1}{2}mv_f^2\ -\ \frac{1}{2}mv_i^2}

Example:












Work can be either positive or negative or zero!
  • If w>0w>0, the net force speeds up the object
  • If w<0w<0, the net force slows down the object

Watch Out!
ΔK ≠12m(Δv)2\Delta K\ \ne\frac{1}{2}m\left(\Delta v\right)^2 , because (Δv)2=(vf−vi)2 ≠vf2  − vi2\left(\Delta v\right)^2=\left(v_f-v_i\right)^2\ \ne v_f^{2\ }\ -\ v_i^2 . If you don't believe me, try plugging in numbers to see!



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Example: Work-Energy Theorem


A 200-kg load is lifted 20.0 m vertically with an acceleration a = 0.160g by a single cable. Determine
(a) the tension in the cable,
(b) the net work done on the load,
(c) the work done by the cable on the load,

(d) the work done by gravity on the load, and
(e) the final speed of the load assuming it started from rest


Solution: Free body diagram


Part a)


To find T
∑Fnet=ma\sum F_{net} = ma
T−mg=maT=ma+mgT-mg =ma\\ T =ma+mg

T=m(a+g)=200(0.160g+g)=200(1.160g)=2273.6N=2.27×103NT = m(a+g) = 200(0.160g+g) = 200(1.160g) = 2273.6N = 2.27\times 10^3N

Part b)


To find net work done on the load

Fnet=0.160mgF_{net} = 0.160mg
Wnet=Fnetdcos(0)°W_{net} = F_{net} d cos (0)\degree
Wnet=(0.160)(200)(9.81)(20)=6.27×103JW_{net} = (0.160)(200)(9.81)(20) = 6.27\times10^3 J

Part c)


work done by cable on the load

Wcable=Tdcos(0)°W_{cable} = T d cos(0)\degree
Wcable=2.27×103×20=4.54×104JW _{cable} = 2.27\times10^3\times 20=4.54\times10^4J

Part d)


work done by gravity on the load
Wgravity=Fgdcos(180)0W_{gravity}=F_gdcos(180)^0
Wgravity=−(200)(9.8)(20)=−3.92×104JW_{gravity} = -(200)(9.8)(20)=-3.92\times10^4J


Part e)


Wnet=ΔKE=(KE)f−(KE)iW_{net} = \Delta KE = (KE)_f-(KE)_i
Wnet=12mvf2−12mvi2W_{net} = \dfrac{1}{2}mv_f^2-\dfrac{1}{2}mv_i^2

since vi=0v_i =0
Wnet=12mvf2W_{net} = \dfrac{1}{2} mv_f^2
vf2=2Wnetmv_f^2 = \dfrac{2W_{net}}{m}
vf=2Wnetm=2×6.27×103200= 7.92 m/sv_f=\sqrt{\frac{2W_{net}}{m}}=\sqrt{\frac{2\times6.27\times10^3}{200}}=\ 7.92\ m/s

A 1.60-m tall person lifts a 2.10-kg book from the ground so it is 2.20 m above the ground. What is the potential energy of the book relative to
(a) the ground
(b) and the top of the person’s head?
(c) How is the work done by the person related to the answers in parts ( a) and ( b)?

A 2.00-kg block is pushed against a spring with negligible mass and force constant k= 400 N/m, compressing it 0.220 m. When the block is released , it moves along a friction-less , horizontal surface and then up a friction-less incline with slope 37.0∘37.0^{\circ}(Fig).

(a) What is the speed of the block as it slides along the horizontal surface after having left the spring?
(b) How far does the block travel up the incline before starting to slide back down?


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A 70kg70kgman carries a 15kg15kgbox up a flight of stairs. If the angle of the stairs is 25°25\degreeand the man travels a total distance of 14m14m, how much work does he do to the box? How much work does he do in total?