Question

When a certain spring is expanded 6 centimeters from its natural position and held fixed, the force necessary to hold it is 1200 dynes. Find the work W required to stretch the spring an additional 6 centimeters

Answer 1

To find the work required to stretch the spring an additional 6 centimeters, we need to consider the potential energy stored in the spring. We can use the formula U = 1/2kx^2 to calculate the potential energy at the initial and final positions and find the difference in potential energy, which represents the work required.

Step-by-step explanation:

To find the work required to stretch the spring an additional 6 centimeters, we need to consider the potential energy stored in the spring. The potential energy stored in a spring can be calculated using the equation U = 1/2kx^2, where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

In this case, the spring is initially stretched 6 centimeters from its natural position, so the initial displacement is 6 centimeters. The work required to stretch the spring an additional 6 centimeters is equal to the difference in potential energy between the final and initial positions.

Since the initial displacement is 6 centimeters, we can use the formula U = 1/2kx^2 to find the potential energy at this position. We are given that the force necessary to hold the spring at this position is 1200 dynes, so we can use the equation F = kx to find the spring constant.

Then, we can calculate the potential energy at the final position by plugging in the values of the spring constant and displacement into the formula U = 1/2kx^2. The work required to stretch the spring an additional 6 centimeters is equal to the difference in potential energy between the final and initial positions.

Answer 2

Answer: The work W required to stretch the spring an additional 6 centimeters

Step-by-step explanation:

The force require to stretch the spring by 6 cm = 1200 dynes =
`1200* 10^(-5) N=0.012 N`

Displacement of from the initial position = 6 cm = 0.06 m

`F=k* x`

`0.012 N=k* 0.06 m`

k = 0.2 N/m

Work done is guiven by:
`(1)/(2)kx^2`

Work done when spring expanded to0.0 6 cm=
`(1)/(2)kx^2=(1)/(2)* 0.2 N/m\time (0.06)^2`..(1)

Work done when spring again expanded to 0.06 cm that is 0.12 m (0.06+0.06 m) =
`(1)/(2)kx^2=(1)/(2)* 0.2 N/m\time (0.12)^2`..(2)

Work done(W) required to stretch the spring an additional 0.06 m: (2)-(1)

`W=(1)/(2)0.2 N/m* (0.12 m)^2-(1)/(2)0.2 N/m*(0.06 m)^2=0.00108 Joules`

The work W required to stretch the spring an additional 6 centimeters