Find position when a particle comes to rest from a potential energy function

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A 1kg particle is in a potential given in joules by U(x)=4x2+5x-3, where x is in meters. The particle is initially at position x=1m and moving with velocity v=6m/s in the +x direction.

Find the position when the particle first comes to rest.

Here is my train of thought

F(x)= -dU/dx

so F(x)= -8x-5

Then i figured since F=ma (mass is 1kg)

that F(x)=a


i used the equation 2a(xf-xi)=v2^2-v1^2 to solve for acceleration which i plugged into above equation using v2=0, v1=6 and xi=1


this gave me
-8x-5= -36/(2x-2)

-16x^2 + 6x + 46=0

then i solved the quadratic, rejected the negative answer since the particle is initially at 1.
My answer is 1.89m which is wrong.

can someone please tell me what i'm doing wrong and point me in the right direction?
 

Answers and Replies

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ehild
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A 1kg particle is in a potential given in joules by U(x)=4x2+5x-3, where x is in meters. The particle is initially at position x=1m and moving with velocity v=6m/s in the +x direction.

Find the position when the particle first comes to rest.

Here is my train of thought

F(x)= -dU/dx

so F(x)= -8x-5

Then i figured since F=ma (mass is 1kg)

that F(x)=a
It is correct up to this point, but wrong afterwards.

i used the equation 2a(xf-xi)=v2^2-v1^2 to solve for acceleration which i plugged into above equation using v2=0, v1=6 and xi=1
The equation you used is valid for constant acceleration only, and it is not the case now.

You would have solved two differential equations, taking into account that the acceleration is the first derivative of the velocity and second derivative of x with respect to time.

But it is not necessary, you have good old conservation of energy, why not use it?

ehild
 

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