A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 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?
 2 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?

This Question is Closed

akash123
 2 years ago
Best ResponseYou've already chosen the best response.1how much work one will do in compressing the spring by 70 mm?

akash123
 2 years ago
Best ResponseYou've already chosen the best response.1or what's the energy stored in the spring when it's compressed by 70 mm?

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0\[E_f={1 \over 2}kx^2={1 \over 2}(1200)(0.07^2)=2.94 [J]\] \[E_i=0\] To find the power it's this equation right? \[P={{\Delta E} \over {\Delta t}}\]

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0\[P={2.940 \over 2}=1.47\] What are the units of Power?

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0Oh right... because it's Energy over time... energy is Joule, time is seconds.. duh. Thanks!

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0To 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?

akash123
 2 years ago
Best ResponseYou've already chosen the best response.1but 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.

akash123
 2 years ago
Best ResponseYou've already chosen the best response.1x will be negative since spring is under compression

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0So then F=kx=1200(0.07)=84 a=F/m=84/10=8.4 ms2

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.

akash123
 2 years ago
Best ResponseYou've already chosen the best response.1no...acceleration of the block wont be same if there is friction..dw:1347418241566:dw

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0Ahhhhh so it would be the net force acting on the block divided by it's mass. Which in this case would be: \[a={F_{net} \over m}={{F_{fric}+F_{spring}} \over m}\]

akash123
 2 years ago
Best ResponseYou've already chosen the best response.1so net force on the block= ( spring force frictional force)

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0Negative frictional force because it's acting in the opposite direction. \[a={{F_{spring}F_{fric}} \over m}={{kx\mu mg} \over m}={{8463.765} \over 10}=2.0235 [ms^{2}]\] That's a big difference.

akash123
 2 years ago
Best ResponseYou've already chosen the best response.1yes...but how did u get the fictional force?

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0\[\mu mg=(0.65)(10)(9.81)=63.765\]

akash123
 2 years ago
Best ResponseYou've already chosen the best response.1ok...you have done correctly.

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0Awesome, Thank you @akash123 This question is complete.
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.