# Hooke’s law – problems and solutions

1. A graph of force (F) versus elongation (x) shown in the figure below. Find the spring constant!

Solution

Hooke’s law formula :

k = F / x

*F = force (Newton)*

*k = spring constant (Newton/meter)*

*x = the change in length (meter) *

Spring constant :

k = 10 / 0.02 = 20 / 0.04

k = 500 N/m

2. Determine the spring constant.

Solution

Spring constant :

k = F / x

k = 5 / 0.01 = 10 / 0.02 = 15 / 0.03 = 20 / 0.04

k = 500 N/m

3. Spring A has the original length of 60 cm and spring B has the original length of 90 cm. Spring A has constant 100 N/m, spring B has constant 200 N/m. The ratio of the change in length of spring A to the change in length of spring B is…

__Known :__

Constant of spring A (k_{A}) = 100 N/m

Constant of spring B (k

_{B}) = 200 N/m

Force on spring A (F_{A}) = F

Force on spring B (F_{B}) = F

__Wanted:__ Δl_{A }: Δl_{B}

__Solution :__

Hooke’s law formula :

Δl = F / k

*Δl** = the change in length, F = force, k = constant *

The change in length of spring A :

Δl_{A} = F_{A} / k_{A} = F / 100

The change in length of spring B :

Δl_{B} = F_{B} / k_{B} = F / 200

The ratio of the change in length of spring A to the change in length of spring B :

Δl_{A} : Δl_{B}

F / 100 : F / 200

1 / 100 : 1 / 200

1 / 1 : 1 / 2

2 : 1

4. A nylon string with original length 20 cm, is pulled by a force of 10 N. The change in length of the string is 2 cm. Determine the magnitude of force if the change in length is 6 cm.

__Known :__

Force (F) = 10 N

The change in length (Δl) = 2 cm = 0.02 m

__Wanted :__ the magnitude of force (F) if Δl = 0.06 m.

__Solution :__

Constant :

k = F / Δl

k = 10 / 0.02 = 500 N/m

The magnitude of force (F) if Δl = 0.06 m :

F = k x

F = (500)(0.06)

F = 30 N

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