## Related questions with answers

Question

Calculate the work required to stretch the following springs 1.25 m from their equilibrium positions. Assume Hooke's law is obeyed. A spring that requires a force of $250 \mathrm{~N}$ to be stretched $0.5 \mathrm{~m}$ from its equilibrium position.

Solution

VerifiedAnswered 10 months ago

Answered 10 months ago

b.

A force of $250$N is required to keep the spring stretched at $x = 0.5$m means by Hooke's law:

$F(0.5) = k(0.5 \text{ m}) = 250 \text{ N}$

Solving for $k$ we get:

$k = 500 \text{ N/m}$

Hence, Hooke's law for this spring is:

$F(x) = 500x$

The spring is compressed from $x = 0$ to $x = 0.4$, so the work is:

$W= \int_0^{1.25} 500x \,dx = \left( 250x^2 \right)\Big \rvert_0^{1.25} = 390.63 \text{ J}$

## Create an account to view solutions

By signing up, you accept Quizlet's Terms of Service and Privacy Policy

## Create an account to view solutions

By signing up, you accept Quizlet's Terms of Service and Privacy Policy

## Recommended textbook solutions

#### Thomas' Calculus

14th Edition•ISBN: 9780134438986 (3 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir10,144 solutions

#### Calculus

3rd Edition•ISBN: 9780134765631 (2 more)Bernard Gillett, Eric Schulz, Lyle Cochran, William L. Briggs14,014 solutions

#### Calculus: Early Transcendentals

8th Edition•ISBN: 9781285741550 (3 more)James Stewart11,085 solutions

#### Calculus: Early Transcendentals

9th Edition•ISBN: 9781337613927 (3 more)Daniel K. Clegg, James Stewart, Saleem Watson11,049 solutions

## More related questions

1/4

1/7