How Does Velocity Change with Position in a Force Field?

[mmattson07]

Homework Statement

The only force acting on a 4.3 kg body as it moves along the positive x-axis has an x component Fx = -8x N, where x is in meters. The velocity of the body at x = 1.6 m is 11 m/s. (a) What is the velocity of the body at x = 4.5 m? (b) At what positive value of x will the body have a velocity of 2.7 m/s?

Homework Equations

F=ma , m=4.3kg, F(x)=-8x
=>v^2=v0^2+2ad

The Attempt at a Solution

a)
=>-8(1.6)/4.3= a = -2.977 m/s^2
=>v^2=v0^2+2ad
=> v^2=(11)^2+2(-2.977)(4.5-1.6)=86.467
=>v=9.299 m/s

b) Here is what I tried:

=>v^2=v0^2+2ad
=>(2.7)^2=(11)^2+2(-2.977)(Xf-1.6)
=>7.29=121-5.954Vf+9.5624
=>Xf=(-113.71+9.5624)/-5.954
=17.492 m

Need to know if this is correct only one submission left.

 

[ehild]

Homework Equations

F=ma , m=4.3kg, F(x)=-8x
=>v^2=v0^2+2ad

v^2=v0^2+2ad is only valid for uniform acceleration, that is, for constant force. That is not the case, the force changes with the position. You can use the work-energy theorem to get the velocity in terms of x, but the work is the integral of the force with respect to x

$$KE_2 -KE_1= \int_{x_1}^{x_2}{Fdx}$$

ehild

 

[mmattson07]

So I just got lucky getting the correct answer for (a) using that equation?

 

[ehild]

Yes. You got a result that is close to the correct answer, v=9.38 m/s, but your answer to b. is far away from the correct one.

ehild

 

[mmattson07]

Could you show how you got v=9.38 m/s for (a)? I know it has to do with E=1/2mv^2 but not getting the right answer.

 

[ehild]

Do you know how to calculate the work done by a non-constant force?

ehild

 

[mmattson07]

It's the integral you posted earlier. I just don't know what to put for the integrand. I tried integrating -8x from 1.6 to 4.5 then setting that equal to 1/2mv^2 but no luck.

 

[ehild]

Remember, the work is equal to the change of kinetic energy.

ehild

 

[mmattson07]

Yes I remember. I know I can find the initial kinetic energy...but to find the final kinetic energy I need to know the total work done, then subtract the initial KE right?

 

[ehild]

Why subtract? KE(final)-KE(initial)=work. ehild