Question

A. Work of 5 Joules is done in stretching a spring from its natural length to 16 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (16 cm)?
b. Work of 5 Joules is done in stretching a spring from its natural length to 12 cm beyond its natural length. What is the spring constant k>0?
c. If the spring is stretched to the same distance (12 cm), what is the restoring force that pulls the spring back toward its equilibrium position?

Answer 1

Answer:

(a) 62.5 N

(b) 694.4 N/m

(c) 41.7 N

Step-by-step explanation:

Work done in stretching a spring is

$W = \frac{1}{2}Fe=\frac{1}{2}ke^2$ since $F=ke$

F is the applied force, E is the elongation and k is the spring constant.

(a) Here, e = 16 cm = 0.16 m

$W = \frac{1}{2}Fe$

$F = \dfrac{2W}{e} = \dfrac{2\times5}{0.16}=62.5\text{ N}$

(b) Here, e = 12 cm = 0.12 m

$W =\frac{1}{2}ke^2$

$k=\dfrac{2W}{e^2}=\dfrac{2\times5}{0.12^2}=694.4\text{ N/m}$

(c) The restoring force is the same as the applied force.

$F = \dfrac{2W}{e} = \dfrac{2\times5}{0.12}=41.7\text{ N}$