Question

the force exerted by a spring is given by the equation

where F is the force (in Newtons), xis the displacement of the spring from the equilibrium position (in centimeters), and k is a spring constant

Create an Excel worksheet showing force as a function of displacement for two springs whose spring constants are 0.1 N/cm and 0.5 N/cm.  Use displacements ranging from 0 to 20 cm (i.e. let x = 0, 1, 2 … 20 cm).

Answer 1

From obey's hooke law, we have

F = k x

where, k = spring constant measured in (N/cm)

F = force exerted by a spring measured in (N)

x = displacement of the spring from an equilibrium position measured in (cm)

At x = 0 cm, we get

F₁ = (0.1 N/cm) (0 cm) 0

F₂ = (0.5 N/cm) (0 cm) 0

At x = 1 cm, we get

F₁ = (0.1 N/cm) (1 cm) 0.1 N

F₂ = (0.5 N/cm) (1 cm) 0.5 N

At x = 2 cm, we get

F₁ = (0.1 N/cm) (2 cm) 0.2 N

F₂ = (0.5 N/cm) (2 cm) 1.0 N

At x = 3 cm, we get

F₁ = (0.1 N/cm) (3 cm) 0.3 N

F₂ = (0.5 N/cm) (3 cm) 1.5 N

At x = 4 cm, we get

F₁ = (0.1 N/cm) (4 cm) 0.4 N

F₂ = (0.5 N/cm) (4 cm) 2.0 N

At x = 5 cm, we get

F₁ = (0.1 N/cm) (5 cm) 0.5 N

F₂ = (0.5 N/cm) (5 cm) 2.5 N

At x = 6 cm, we get

F₁ = (0.1 N/cm) (6 cm) 0.6 N

F₂ = (0.5 N/cm) (6 cm) 3.0 N

At x = 7 cm, we get

F₁ = (0.1 N/cm) (7 cm) 0.7 N

F₂ = (0.5 N/cm) (7 cm) 3.5 N

At x = 8 cm, we get

F₁ = (0.1 N/cm) (8 cm) 0.8 N

F₂ = (0.5 N/cm) (8 cm) 4.0 N

At x = 9 cm, we get

F₁ = (0.1 N/cm) (9 cm) 0.9 N

F₂ = (0.5 N/cm) (9 cm) 4.5 N

At x = 10 cm, we get

F₁ = (0.1 N/cm) (10 cm) 1.0 N

F₂ = (0.5 N/cm) (10 cm) 5.0 N

At x = 11 cm, we get

F₁ = (0.1 N/cm) (11 cm) 1.1 N

F₂ = (0.5 N/cm) (11 cm) 5.5 N

At x = 12 cm, we get

F₁ = (0.1 N/cm) (12 cm) 1.2 N

F₂ = (0.5 N/cm) (12 cm) 6.0 N

At x = 13 cm, we get

F₁ = (0.1 N/cm) (13 cm) 1.3 N

F₂ = (0.5 N/cm) (13 cm) 6.5 N

At x = 14 cm, we get

F₁ = (0.1 N/cm) (14 cm) 1.4 N

F₂ = (0.5 N/cm) (14 cm) 7.0 N

At x = 15 cm, we get

F₁ = (0.1 N/cm) (15 cm) 1.5 N

F₂ = (0.5 N/cm) (15 cm) 7.5 N

At x = 16 cm, we get

F₁ = (0.1 N/cm) (16 cm) 1.6 N

F₂ = (0.5 N/cm) (16 cm) 8.0 N

At x = 17 cm, we get

F₁ = (0.1 N/cm) (17 cm) 1.7 N

F₂ = (0.5 N/cm) (17 cm) 8.5 N

At x = 18 cm, we get

F₁ = (0.1 N/cm) (18 cm) 1.8 N

F₂ = (0.5 N/cm) (18 cm) 9.0 N

At x = 19 cm, we get

F₁ = (0.1 N/cm) (19 cm) 1.9 N

F₂ = (0.5 N/cm) (19 cm) 9.5 N

At x = 20 cm, we get

F₁ = (0.1 N/cm) (20 cm) 2.0 N

F₂ = (0.5 N/cm) (20 cm) 10 N