Question

A horizontal spring with constant k=1200 and mass 10kg attached to the end. The mass is on a horizontal frictionless surface.
Question: How much power is required to compress the spring 70mm in a period of 2 seconds?

Answer 1

how much work one will do in compressing the spring by 70 mm?

Answer 2

or what's the energy stored in the spring when it's compressed by 70 mm?

Answer 3

\[E_f={1 \over 2}kx^2={1 \over 2}(1200)(0.07^2)=2.94 [J]\]
\[E_i=0\]
To find the power it's this equation right?
\[P={{\Delta E} \over {\Delta t}}\]

Answer 4

yes.....good:)

Answer 5

\[P={2.94-0 \over 2}=1.47\]
What are the units of Power?

Answer 6

watt or joule/s

Answer 7

Oh right... because it's Energy over time... energy is Joule, time is seconds.. duh.
Thanks!

Answer 8

yes...:)

Answer 9

To find the initial acceleration if the spring was released do i use F=ma?
Calculate the Force using Hooke's Law \[F=-kx\] and plug into \[a={F \over m}\]
Right? But is x negative as it's compressed by 70mm or is it positive?

Answer 10

but we make some assumptions in this question i.e. we are compressing the spring smoothly(slowly) such that there is no change in kinetic energy of mass attached with the spring.

Answer 11

x will be negative since spring is under compression

Answer 12

So then F=-kx=-1200(-0.07)=84
a=F/m=84/10=8.4 ms-2

Answer 13

I think they're trying to trick us. Because if that's right above it doesn't factor in friction, and the next question asks what if the surface has friction of 0.65 what is the initial acceleration.
Well it would be the same answer.

Answer 14

no...acceleration of the block wont be same if there is friction..