Learn How to Solve a Difficult Homework Question on Hooke's Law

In summary: Initial energy stored in the spring is equal to the final kinetic energy of the trolley.In summary, the conversation is about a homework question involving a trolley on a frictionless air track with a spring attached to one end. The initial acceleration of the trolley, initial energy stored in the spring, and final speed of the trolley are all calculated using equations such as F = ma and W = \frac{1}{2}kx^2. The conversation also discusses the transformation of energy from the potential energy of the spring to the kinetic energy of the trolley.
  • #1
charikaar
5
0
Hello,

I found one of my this homework question difficult and was wondering if anyone could please help me.

A 180g trolley is placed on frictionless air track. One end of the trolley is attached to a spring of spring constant 50 N/m. The trolley is pushed against a fixed support so that the compression of the spring is 8 cm. The trolley is then released.

(a) What is the initial acceleration of the trolley when it is released?
(b) What is the initial energy stored in the spring?
(c) Calculate the final speed of the trolley along the frictionless track. You may assume that there is 100% transfer of energy from the spring to the trolley.

Any help would greatly be appreciated.



charikaar
 
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  • #2
(a) F = ma
(b) [tex]W = \frac{1}{2}kx^2[/tex] You can obtain that by integrating, but I'm not sure if you're familiar with that stuff.
(c) Energy principle
 
  • #3
hello,

Thanks for the quick reply.


Mass of trolley: 180g or 0.18 kg

compression: 8cm or 0.08m

Using F=extentionxconstant to find Force
50/0.08=625N

Now using F=ma to find acceleration

625/0.18=3472.22...Am i correct?

(b) using W=1/2kx^2

25x0.64=16J

(c) I've no idea how to start this one.


thank you.


Charikaar
 
  • #4
(a) Instead of calculating the force right in the beginning, solve the problem with the letters, and into the final equation, plug in the requested stuff. Oh, and F = -kx, not -k/x
(b) Are you using SI-units? I got a different answer.
(c) What sort of energy does the potential energy of the spring transform into?
 
Last edited:
  • #5
Hi again,

I've just started AS physics and am not familiar with this stuff much. I don't know where does my teacher get these question from when they are not in the book.


(a) k=50 N/m, x=0.08 m using F=kx, F=50x0.08=4N

Now using F=ma to acceleration
a=F/m, 4/0.18=22.22...N/mkg


(b) This is AS Physics question.
Elastic Potential Energy=1/2xstretching forcexextension
=1/2x8x0.08=0.32J




(c)I think it transfers to energy of movement. Could you please give me equation for the solution.


Thanks for your help.


Charikaar
 
  • #6
(a) Correct. From F = ma, we can see that the unit of F is [itex]kgm/s^2[/itex] This divided by kg, we get [itex]m/s^2[/itex], which is the unit of acceleration.
(b) My answer's half that.
(c) Kinetic energy: [tex]\frac{1}{2}mv^2[/tex].
 

FAQ: Learn How to Solve a Difficult Homework Question on Hooke's Law

What is Hooke's Law?

Hooke's Law is a principle in physics that describes the relationship between the force applied to an elastic material and the resulting deformation or change in shape of the material.

Who discovered Hooke's Law?

Hooke's Law was discovered by English scientist Robert Hooke in the 17th century.

What is the equation for Hooke's Law?

The equation for Hooke's Law is F = -kx, where F is the force applied, k is the spring constant, and x is the displacement of the material from its original position.

What is the significance of Hooke's Law?

Hooke's Law is significant because it helps us understand the behavior of elastic materials and how they respond to external forces. It is also used in various fields of science and engineering, such as mechanics and materials science.

Are there any limitations to Hooke's Law?

Yes, there are limitations to Hooke's Law. It only applies to materials that exhibit elastic behavior, meaning they return to their original shape after the applied force is removed. It also assumes that the force applied is within the material's elastic limit, beyond which the material will be permanently deformed.

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