Principles of Physics - Foederer
Energy is stored in a spring when work is done
to compress or elongate it
Compression or elongation = change in length
= x (m)
Springs have a spring constant (k) that tells
how much force is required to change the
length of a spring by 1m.
Robert Hooke put it all together→ Fs = kx
Fs = spring force (N)
k
= spring constant (N/m)
x
= change in length (m)
Based on Hooke’s experiments, it was
determined that the energy stored in the
spring can be calculated using the following
formula:
PEs = ½ kx2
PEs = elastic potential energy (joules)
k
= spring constant (N/m)
x
= change in length (m)
A spring (k = 340 N/m) is compressed 0.02m.
How much energy is stored in the spring?
PEs = ½ kx2
= ½(340 N/m)(0.02m)2
= 0.068 J
A 5 kg block sliding on a frictionless horizontal
surface with a speed of 5 m/s collides with a
spring (k = 120 N/m) and comes to rest. How
much will the spring be compressed?
KE→PEs
½ mv2 = ½ kx2
½ (5kg)(5 m/s)2 = ½(120 N/m)(x)2
x = 1.02 m
The diagram at right shows a 0.1kilogram apple attached to a branch of
a tree 2 meters above a spring on the
ground below.
The apple falls and hits the spring,
compressing it 0.1 meter from its rest
position. Assume all energy is conserved
within the system.
1.
Write the energy transformations that occur
as the apple falls from the tree.
PE
KE
PEs
______→
______→
______
The apple (0.1 kg) falls 2m and hits the
spring, compressing it 0.1 meter from
its rest position. Assume all energy is
conserved within the system.
2. Determine the gravitational PE of the
apple before it falls.
PE= mgh
PE = 0.1(9.8)(2)
PE = 1.96 J
The apple (0.1 kg) falls and hits the
spring, compressing it 0.1 meter from
its rest position. Assume all energy is
conserved within the system.
3. Determine the kinetic energy and the
velocity of the apple when it hits the
spring.
KE = ½ mv2
1.96 = ½ (0.1)(v2)
½ (0.1)
½ (0.1)
√ 39.2 = √ v2
v = 6.26 m/s
The apple (0.1 kg) falls and hits the
spring, compressing it 0.1 meter from
its rest position. Assume all energy is
conserved within the system.
4. Determine the elastic potential
energy and the spring constant of
this spring?
PEs = ½ kx2
1.96
= ½(k)(0.1)2
½ (0.1)2
½ (0.1)2
k = 392 N/m