A 31.0 cm long spring is hung vertically from a ceiling and stretches to 36.9 cm when an 8.00 kg mass is hung from its free end. (a) Find the spring constant (in N/m).
 

(b) Find the length of the spring (in cm) if the 8.00 kg weight is replaced with a 185 N weight.

The length of the spring is 44.92 cm

Explanation:

Hooke's Law

Suppose a spring of constant k and natural length x0. If a force F is applied to the spring and it stretches to a distance x1. Hooke's Law states that:

Where x is the elongation of the spring:

We are given the characteristics of a spring of x0=31 cm. When a mass of m=8 Kg is hung from the spring, it stretches to x1=36.9 cm. We need to calculate the force of the mass of 8 Kg. It can be done by calculating the weight:

The elongation of the spring is

Converting to meters:

(a)

From Hooke's Law, we solve for k:

(b) With the value of k, the equation for the spring is:

Now if a weight of F=185 N is hung from the spring, the elongation is:

Thus, the length of the spring is:

The length of the spring is 44.92 cm

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j