Name
Date
Pd
Unit 6: Work and Energy
Hooke’s Law and EPE Worksheet
Suppose in the lab one group found that F=1000. mN (x) . Construct a graphical representation of
force vs. displacement. (Hint: make the maximum displacement 0.25 m.)
1. Graphically determine the amount of energy stored while stretching the spring described above
from x = 0 to x = 10. cm.
250
150
Wspr = area
= ½(Fspr)(x)
= ½(kx)(x)
= ½ kx2
= ½(100)(0.1)=5J
50
0
0.05
0.15
0.25
x (m)
2. Graphically determine the amount of energy stored while stretching the spring described above from
x = 15 to x = 25 cm.
Wspr = area = ½(Fspr)(x)
= 20J
250
Area = ½ (0.1)(100)=5J
150
Area = (0.1)(150)=15J
50
0
0.05
0.15
x (m)
0.25
The graph at left was made from data collected during an investigation of the relationship between the
amount two different springs stretched, when different forces were applied.
36.0
3. For each spring determine the spring constant.
Data Set 1
Data Set 2
32.0
Spring constant, k, is the slope of the F-x plot
Data set 1: k = 8 N/m
Data set 2: k = 5 N/m
28.0
24.0
4. For each spring, compare
a. the amount of force required to stretch the spring 3.0
m. F=kx
Data set 1: F3 = 24 N
Data set 2: F3 = 15 N
Force
(N)
20.0
16.0
12.0
8.00
4.00
0.00
0.00
2.00
4.00
x (m)
6.00
8.0
0
b. the US stored in each spring when stretched 3.0 m.
Stored Us is equal to Work done against the spring (-Ws)
to stretch the spring from its equilibrium to x
Data set 1: -Ws=Us = -Ws= 1/2 kx2=1/2 (8)(32) = 36 J
Data set 2: -Ws=Us = -Ws= 1/2 kx2=1/2 (3)(32) = 13.5J
5. Determine the amount that spring 2 needs to be stretched in order to store 24 joules of energy.
 Wspr  U S  12 k (x 2 )
24  12 5x 2
x  3.10m
6. You have two springs that are identical except that spring 1 is stiffer than spring 2 (k1>k2). On which
spring is more work done
a) if they are stretched using the same force? the less stiff spring, spring 2
b) if they are stretched the same distance? the stiffer spring, spring1
7. The spring below has a spring constant of 10. N/m. If the block is pulled 0.30 m horizontally to
the right, and held motionless, what force does the spring exert on the block? Sketch a force
diagram for the mass as you hold it still. (Assume a frictionless surface.)
FN
Fspr
Fpull
Fg
Fspr  kx  10(0.3)  3N
8. The spring below has a spring constant of 20. N/m. The s between the box and the surface is 0.40.
a. The box is pushed to the right, then released. Once released, the box remains in place, in other
words, it doesnt return to the equilibrium position where the force of the spring is zero. Draw a
force diagram for the box above when the spring is stretched, yet the box is stationary.
FN
Fspr
fs
//
//
1 kg
Fg
b. What is the maximum distance that the spring can be stretched
from equilibrium before the box begins to slide back?
F
0
 Fspr  fs  0
Fspr  fs
kx   s FN
20x  0.4(mg )
x  0.196m
x
9. a) A spring is stretched from a rest length of 25 cm to a length of 30 cm. A certain amount of energy,
US1, is stored. The spring is then stretched from 30 to 35 cm. How does the new amount of
energy stored compare to US1?
(-WS=US so US=1/2kx2)
Work done against spring changes or stores spring potential energy:
-WS= ½k(x2) = US where Dx is the displacement from equilibrium.
Stored spring potential energy or Us depends on the square of springs displacement from
equilibrium so as the spring is stretched out more, the energy stored increases quadratically.
Therefore the energy stored by stretching the spring 5cm from 25cm to 30 cm is LESS than the
energy stored going an additional 5cm from 30 to 35cm.
In this problem the equilibrium is at 25cm so as the spring is stretched from 25-30 cm, x=5cm=0.05m
 Ws  12 k (x 2 )
W2530cm  12 k (0.052 )  12 k (0.0025)
Energy stored from 30-35 is W going from equilibrium (25cm) to 35 minus work going from equilib to 30
W3035cm  W2535cm  W2530cm  12 k (.12 )  12 k (.052 )  12 k (0.01  .0025)  12 k (.0075)
Stores 3 xs more energy in 5cm stretch from 30-35cm than the 5cm stretch from 25-30cm
b) A block is raised from the ground to a height H and now contains a certain amount of energy
Ug1. The box is then raised to a height of 2H. How does the new amount of energy stored
compare to Ug1?
(-Wg=Ug so Ug=mgh)
Ug = mgy. It depends linearly on vertical displacement or height. Therefore if a box is raised
to double the height, it will store double the gravitational potential energy.
c) An object is moving with speed v and contains a certain amount of energy K1. The speed is
increased to 2v. How does the new amount of energy stored compare to K1?
K = ½mv2 . K depends on the square of v. Therefore if v is doubled, K increases by 4