Home > Physics > Work

If a spring has a constant of #3 (kg)/s^2#, how much work will it take to extend the spring by #37 cm #?

Answer 1

Sebastian Acosta

Approximately #0.07# joules.

On a spring, work is provided by:

#W=1/2kx^2#

where:

#k# is the spring constant, usually in newtons per meter

#x# is the extension of the spring, usually in meters

So here: #x=37 \ "cm"=0.37 \ "m"#.

And so,

#W=1/2*3 \ "kg s"^-2*(0.37 \ "m")^2#

#=1/2*3 \ "kg s"^-2*0.1369 \ "m"^2#

#=1/2*3*0.1369 \ "J" \ (because 1 \ "J"-=1 \ "kg m"^2 \ "s"^-2)#

#=0.06845 \ "J"#

#~~0.07 \ "J"#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Nora Arzate

To calculate the work done in extending the spring, you can use the formula for the potential energy stored in a spring: ( W = \frac{1}{2}kx^2 ). Given ( k = 3 , \text{(kg/s}^2) ) and ( x = 0.37 , \text{m} ), plug these values into the formula:

[ W = \frac{1}{2} \times 3 \times (0.37)^2 ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.