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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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akash123Best ResponseYou've already chosen the best response.1
how much work one will do in compressing the spring by 70 mm?
 one year ago

akash123Best ResponseYou've already chosen the best response.1
or what's the energy stored in the spring when it's compressed by 70 mm?
 one year ago

PhoenixFireBest 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? \[P={{\Delta E} \over {\Delta t}}\]
 one year ago

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

PhoenixFireBest ResponseYou've already chosen the best response.0
Oh right... because it's Energy over time... energy is Joule, time is seconds.. duh. Thanks!
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
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?
 one year ago

akash123Best ResponseYou've already chosen the best response.1
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.
 one year ago

akash123Best ResponseYou've already chosen the best response.1
x will be negative since spring is under compression
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
So then F=kx=1200(0.07)=84 a=F/m=84/10=8.4 ms2
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
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.
 one year ago

akash123Best ResponseYou've already chosen the best response.1
no...acceleration of the block wont be same if there is friction..dw:1347418241566:dw
 one year ago

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

akash123Best ResponseYou've already chosen the best response.1
so net force on the block= ( spring force frictional force)
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
Negative 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.
 one year ago

akash123Best ResponseYou've already chosen the best response.1
yes...but how did u get the fictional force?
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
\[\mu mg=(0.65)(10)(9.81)=63.765\]
 one year ago

akash123Best ResponseYou've already chosen the best response.1
ok...you have done correctly.
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
Awesome, Thank you @akash123 This question is complete.
 one year ago
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