Question

1.A hand exerciser utilizes a coiled spring. A force of 89.0 N is required to compress the spring by 0.0191 m. Determine the force needed to compress the spring by 0.0508 m.

Answer 1

Answer:

236.71 N

Explanation:

Applying,

F = ke................. Equation 1

Where F = force, k = spring constant, e = extension.

make k the subject of the equation

k = F/e.............. Equation 2

From the question,

Given: F = 89 N, e = 0.0191 m

Susbtitute these values into equation 2

k = 89/0.0191

k = 4659.69 N/m

When e = 0.0508 m

The force needed is given as

F = (4659.69×0.0508)

F = 236.71 N

Answer 2

Answer:

Force needed to compress spring by 0.0508 m = 236.71 N (Approx.)

Explanation:

Given:

Force applied = 89 N

Length of spring compress = 0.0191 m

Find:

Force needed to compress spring by 0.0508 m

Computation:

Using Hooke's law

F = K(x)

Constant K = F / x

Constant K = 89 / 0.0191

Constant K = 4,659.68 N/m

Force needed to compress spring by 0.0508 m

So,

Force needed to compress spring by 0.0508 m = K(X)

Force needed to compress spring by 0.0508 m = (4,659.68)(0.0508)

Force needed to compress spring by 0.0508 m = 236.71 N (Approx.)