Question

A spring of spring constant 5 N/m and natural length
7 cm is stretched from 8 cm to 9 cm of length.
What is the work done by the spring force?
(1) 7.5*10-4 J (2) 5 x 10-4 J
(3) -7.5 x 10-4 J (4) 25*10-4 J​

Answer 1

Hello Dear,

◆ Answer - (1)

W = 7.5×10^-4 J

◆ Explanation -

# Given -

k = 5 N/m

x = 7 cm = 0.07 m

x1 = 8 cm = 0.08 m

x2 = 9 cm = 0.09 m

# Solution -

Work done in stretching wire from its original length to 8 cm is -

W1 = k(x1-x)²/2

W1 = 5 × (0.08-0.07)² / 2

W1 = 2.5×10^-4 J

Work done in stretching wire from its original length to 9 cm is -

W2 = k(x2-x)²/2

W2 = 5 × (0.09-0.07)² / 2

W2 = 10×10^-4 J

So work done from stretching spring from 8 cm to 9 cm is -

W = W2 - W1

W = 10×10^-4 - 2.5×10^-4

W = 7.5×10^-4 J

Therefore, work done in stretching spring is 7.5×10^-4 J.

Thanks dear...

Answer 2

Answer:

A spring with a force constant of 5 N/m and a natural length  of 7 cm is stretched from 8 cm to 9 cm of length, the work done by the spring force is $7.5*10^{-4}J$

Explanation:

• The elastic potential energy is the energy stored when a force applied to deform an elastic object

• A spring with elastic constant K compressed or expanded at a distance x from its mean position, then the potential energy stored is

        $PE=\frac{1}{2}Kx^{2}$

• The work done by the spring is equal to the change in stored energy

        $W=\frac{1}{2}Kx^{2}$

From the question,

the spring constant $K=5N/m$

the natural length of the spring $l=7cm=0.07m$

work done by the spring when it is stretched to $8cm$from $7cm$

$W_{1}=\frac{1}{2}(5(0.08-0.07)^{2} )=2.5*10^{-4}J$

work done by the spring when it is stretched to $9cm$ from $7cm$

$W_{2}=\frac{1}{2}(5(0.09-0.07)^{2} )=10*10^{-4}J$

work done by the spring when it stretched from $8cm$ to $9cm$  

$W_{2} -W_{1}=(10-2.5)*10^{-4}J=7.5*10^{-4}J$

option A is the right answer