Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (19 cm) ? Don't forget to enter the correct units.

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F. dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k (0.0361-0)

5*2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 * 0.19

F = 52.63 N

The applied force is 52.63N