Hooke's Law Problem Homework: Extension & Work Done

[Koncept]

Homework Statement

A mass of 1.0kg is attached to a spring obeying Hooke's Law F = k.s, where F is the force applied and s the spring extension. The spring constant, k is 50 N/m. The spring and the object lie on a surface tilted 45 degrees with respect to the vertical Neglect friction and answer the following questions:

a. What is the extension of the spring?

b. What is the work done by gravity in extending the spring by the above extension?

Homework Equations

F = k.s

W = f.d

The Attempt at a Solution

a)

1kg = 9.8N (force)

Then resolve force into vert and horiz components= Tan45 x 9.8 = 9.8 (as tan45 is 1).

Rearrange F = k.s into F/k = s

9.8/50 = 0.196m = s

b)

Work done by gravity:

9.8N x 0.196 = 1.92 Joules

Does anyone know if this is right?

Thanks!

 

[JaWiB]

Then resolve force into vert and horiz components= Tan45 x 9.8 = 9.8 (as tan45 is 1).

This is wrong. Draw the right triangle representing the force and its components; the hypotenuse is the weight force (directed straight downward), so the components must be less than that.

 

[ideasrule]

Homework Statement

a)

1kg = 9.8N (force)

Then resolve force into vert and horiz components= Tan45 x 9.8 = 9.8 (as tan45 is 1).

Rearrange F = k.s into F/k = s

9.8/50 = 0.196m = s

Where did you get tan45 from? The parallel component of the weight is 9.8*sin(45).

b)

Work done by gravity:

9.8N x 0.196 = 1.92 Joules

Does anyone know if this is right?

Thanks!

No, because W=F*d only works if both F and d remain constant. F certainly does, but not d. Use the formula for potential energy instead: U=(1/2)kx^2. That's equal to the work done by gravity.

 

[Koncept]

Ahh yes that makes a lot more sense! Thanks for the help :)

So I now make that:

a) 0.139m

b) 0.483 J