Chapter 4.6 – Elastic Potential Energy and Simple Harmonic Motion
Page 195, Practice Questions
1.
Spring Forces Questions
a)
GRASS METHOD PROCESS
RATIONALE
G: m = 0.65 kg; ∆x = 0.44m;
g = 9.8 m/s 2
R: k
1) The force of gravity on the mass is directed
A: ⃗⃗⃗⃗
Fg = mg [down] = down. The restorative spring force on the mass
mg [up];
points up because the spring is stretched down.
To calculate the total force, subtract the
⃗⃗⃗⃗
Fx = −k∆x⃗ = k∆x⃗ [up]
magnitudes.
2) Note: Recognize that the restorative force on
the spring, k∆x⃗, is positive because it acts
opposite to the force acting on it
S:
3) Since the mass is not accelerating (at rest), all of
the forces are at equilibrium. So according to
⃗ =0
Newton’s 2nd law, ∑ F
1) Set the sum of forces to equal zero, as the spring
and object is in equilibrium
⃗ =0
∑F
Fx + Fg = 0
2) The forces acting on the spring are the force
from the mass and the force of gravity acting in
the opposite direction; where up is positive and
down is negative
k∆x⃗ − mg = 0
So, k =
=
mg
∆x
m
(0.65 kg)(9.8 2 )
s
0.44m
k = 14.5 N/m
S: The spring constant is 14 N/m
3) Rearrange the equation to solve for k
b)
GRASS METHOD PROCESS
G: k = 14.5 N/m; ∆x = 0.74m;
g = 9.8 m/s 2
R: m
A: k∆x⃗ − mg = 0
S: k∆x⃗ − mg = 0
k∆x⃗ = mg
So, m =
=
k∆x
g
(14.5 N/m)(0.74 m)
9.8 m/s2
m = 1.1 kg
S: The new mass is 1.1 kg.
RATIONALE
This equation where the sum of the
forces is useful because there is no
acceleration of the mass and the forces
acting on it are in equilibrium
Solve for m
2.
GRASS METHOD PROCESS
N
G: m = 5.3 kg; k = 720 ;
m
g = 9.8 m/s 2 ; ∆x = 0.36 m
R: ⃗FNet ; a⃗
A: ⃗FNet = ⃗Fx + ⃗Fg
Where, ⃗Fg = mg [down];
⃗Fx = −k∆x⃗ = k∆x⃗ [down]
RATIONALE
- The force of gravity on the
mass points down. The
spring force on the mass
points down because the
spring is compressed
upward
S: ⃗FNet = ⃗Fx + ⃗Fg
⃗FNet = mg [down] + k∆x [down]
9.8m
N
= (5.3 kg) ( 2 ) [down] + 720 ( ) (0.36m)[down]
s
m
𝐅𝐍𝐞𝐭 = 𝟑𝟏𝟏𝐍
So, ⃗FNet = ma⃗
⃗⃗⃗⃗⃗⃗⃗⃗
FNet
311 N [down]
a⃗ =
=
m
5.3 kg
𝐦
𝐚⃗ = 𝟓𝟗 𝟐 [𝐝𝐨𝐰𝐧]
𝐬
S: The force on the mass is 310 N [down], and the
acceleration is 59 m/s 2 [down]
In a FBD,
x = 0.36 m
-kx
x=0
mg
Then use the equation,
⃗FNet = ma⃗ to