Problem 5: We are designing a delivery ramp for crates containing exercise equipment. The 1470N crates will move at 1.8 m/s at the top of a ramp that slopes downward at 22 degrees. The ramp exerts a 550N kinetic friction force on each crate, and the maximum static friction also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 8.0 meters along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the maximum force constant of the spring which will meet the design criteria. Take sin22° = 0.374 and g = 10
m
/
s
2
m
/
s
2
.
This problem is adapted from the question mailed by a student.
Solution : Here the critical part of the question is to understand that the force required to return the crate up is equal to the sum of the component of its weight down the incline and the maximum static friction so that block can be initiated up the incline. Hence, spring force for this condition is :
F
S
=
m
g
sin
θ
+
F
K
=
m
g
sin
22
0
+
550
=
1470
X
0.374
+
550
=
1100
N
F
S
=
m
g
sin
θ
+
F
K
=
m
g
sin
22
0
+
550
=
1470
X
0.374
+
550
=
1100
N
Let “x” be the compression in the spring. Then, applying Hooke’ law, we get the compression in the spring as :
F
S
=
K
x
=
1100
F
S
=
K
x
=
1100
The system here is an isolated system with both conservative and nonconservative force. Hence,
W
F
=
Δ
K
+
Δ
U
W
F
=
Δ
K
+
Δ
U
Here,
⇒
W
F
=

550
x
8
=

4400
J
⇒
W
F
=

550
x
8
=

4400
J
⇒
Δ
K
=
K
f
−
K
i
=
0
−
1
2
X
147
X
1.82
=

238.14
J
⇒
Δ
K
=
K
f
−
K
i
=
0
−
1
2
X
147
X
1.82
=

238.14
J
⇒
Δ
U
=
1
2
k
x
2
−
m
g
h
=
1
2
k
x
2
−
1470
X
8
X
sin
22
0
⇒
Δ
U
=
1
2
k
x
2
−
m
g
h
=
1
2
k
x
2
−
1470
X
8
X
sin
22
0
⇒
Δ
U
=
1
2
k
x
2
−
1470
X
8
X
sin
22
0
=
1
2
k
x
2
−
4398
⇒
Δ
U
=
1
2
k
x
2
−
1470
X
8
X
sin
22
0
=
1
2
k
x
2
−
4398
Putting values in the equation, we have :
⇒
W
F
=
Δ
K
+
Δ
U
⇒
W
F
=
Δ
K
+
Δ
U
⇒

4400
=

238
+
1
2
k
x
2
−
4398
⇒

4400
=

238
+
1
2
k
x
2
−
4398
⇒
1
2
k
x
2
=
238
⇒
1
2
k
x
2
=
238
⇒
k
x
2
=
476
⇒
k
x
2
=
476
Combining with Hooke’s equation,
⇒
K
=
476
K
2
1100
2
⇒
K
=
476
K
2
1100
2
⇒
K
=
1100
2
476
=
2542
N
/
m
⇒
K
=
1100
2
476
=
2542
N
/
m
We can see that a spring having spring constant less than this value will also not return the crate. Hence, this spring constant represents the maximum value of the spring constant, which meets the required design criterion.