Question

Suppose that 5 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 51 cm. (a) How much work is needed to stretch the spring from 41 cm to 46 cm? (Round your answer to two decimal places.) 5 Incorrect: Your answer is incorrect. J (b) How far beyond its natural length will a force of 25 N keep the spring stretched? (Round your answer one decimal place.) cm

Answer 1

a) 0.6 joules of work are needed to stretch the spring from 41 centimeters to 46 centimeters.

b) The spring must be 5.6 centimeters far from its natural length.

Explanation:

a) The work done to stretch the ideal spring from its natural length is defined by the following definition:

`W = (1)/(2)\cdot k\cdot (x_(f)-x_(o))^(2)` (1)

Where:

`k` - Spring constant, measured in newtons.

`x_(o)`,
`x_(f)` - Initial and final lengths of the spring, measured in meters.

`W` - Work, measured in joules.

The spring constant is: (
`W = 5\,J`,
`x_(o) = 0.36\,m`,
`x_(f) = 0.51\,m`)

`k = (2\cdot W)/((x_(f)-x_(o))^(2))`

`k = (2\cdot (5\,J))/((0.51\,m-0.36\,m)^(2))`

`k = 444.44\,(N)/(m)`

If we know that
`k = 444.44\,(N)/(m)`,
`x_(o) = 0.41\,m` and
`x_(f) = 0.46\,m`, then the work needed is:

`W = (1)/(2)\cdot \left(444.44\,(N)/(m) \right)\cdot (0.46\,m-0.41\,m)^(2)`

`W = 0.555\,J`

0.6 joules of work are needed to stretch the spring from 41 centimeters to 46 centimeters.

b) The elastic force of the ideal spring (
`F`), measured in newtons, is defined by the following formula:

`F = k\cdot \Delta x` (2)

Where
`\Delta x` is the linear difference from natural length, measured in meters.

If we know that
`k = 444.44\,(N)/(m)` and
`F = 25\,N`, then the linear difference is:

`\Delta x = (F)/(k)`

`\Delta x = (25\,N)/(444.44\,(N)/(m) )`

`\Delta x = 0.056\,m`

The spring must be 5.6 centimeters far from its natural length.