## PhoenixFire 3 years ago 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?

1. akash123

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

2. akash123

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

3. PhoenixFire

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

4. akash123

yes.....good:)

5. PhoenixFire

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

6. akash123

watt or joule/s

7. PhoenixFire

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

8. akash123

yes...:)

9. PhoenixFire

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?

10. akash123

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.

11. akash123

x will be negative since spring is under compression

12. PhoenixFire

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

13. PhoenixFire

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.

14. akash123

no...acceleration of the block wont be same if there is friction..|dw:1347418241566:dw|

15. PhoenixFire

Ahhhhh so it would be the net force acting on the block divided by it's mass. Which in this case would be: $a={F_{net} \over m}={{F_{fric}+F_{spring}} \over m}$

16. akash123

so net force on the block= ( spring force- frictional force)

17. PhoenixFire

Negative frictional force because it's acting in the opposite direction. $a={{F_{spring}-F_{fric}} \over m}={{-kx-\mu mg} \over m}={{84-63.765} \over 10}=2.0235 [ms^{-2}]$ That's a big difference.

18. akash123

yes...but how did u get the fictional force?

19. PhoenixFire

$\mu mg=(0.65)(10)(9.81)=63.765$

20. akash123

ok...you have done correctly.

21. PhoenixFire

Awesome, Thank you @akash123 This question is complete.

22. akash123

:)