How Do You Calculate Work Done by a Variable Force Along a Displacement?

[Sneakatone]

A particle moves along the x-axis from x=3 to x=5 m. A force F_x(x)=2x^2+8x acts on the particle (the distance x is measured in meters and the force in Newtons). calculate the work done by the force F_x(x) during this motion.

I tried using
[f(5)-f(3)]*[5-3]
but i don't know what I am suppose to do with the equation.

 

[kinof]

Since the force is varying, you need to find the work along each differential portion of distance between x=3 and x=5. Think about the definition of work, and apply that.

 

[Sneakatone]

I did f(5)=90N*5m=450J
f(3)=42*3m=126J

is that right?

 

[haruspex]

No, you need to use integration. E = ∫F.dx

 

[Sneakatone]

you mean find the derivative?
4x+8

 

[haruspex]

No, I don't mean find the derivative. I mean exactly what I wrote: take the integral of the force wrt x. That's the definition of work done.

 

[Sneakatone]

so anti derivative..
2/3x^3+4x^2+C

 

[haruspex]

so anti derivative..
2/3x^3+4x^2+C

Yes. Plug in the limits for x.

 

[Sneakatone]

do you want f(5)=2/3(5)^3+4(5)^2=183.33
f(3)=2/3(3)^3+4(3)^2= 54
or
f(2)=2/3(2)^3+4(2)^2=21.33
??

 

[haruspex]

do you want f(5)=2/3(5)^3+4(5)^2=183.33
f(3)=2/3(3)^3+4(3)^2= 54
or
f(2)=2/3(2)^3+4(2)^2=21.33
??

Do you not know how to perform a definite integral? You evaluate the indefinite integral at each end of the range and take the difference (end minus start). The constant of integration is the same for both so cancels out.

 

[Sneakatone]

I don't really know how to do it from here.

 

[iRaid]

You do know this right?:
$$W=\int_a^b F(x)dx$$
Therefore:
$$W=\int_3^5 (2x^{2}+8x)dx$$
Can you solve that?

Edit:
Looking above, you don't know how to evaluate a definite integral:
Find the integral from above and pug in the top limit of integration (5) - the bottom limit of integration (3).
Take some function f(x) and find the anti derivative, F(x).
$$F(x)|_a^b = F(a)-F(b)$$

 

[haruspex]

$$F(x)|_a^b = F(a)-F(b)$$

I think you mean $$F(x)|_a^b = F(b)-F(a)$$

 

[Sneakatone]

i ended up with 129.3 J ,
Thanks for the help!

 

[haruspex]

I don't really know how to do it from here.

Using your notation in post #9, it's f(5) - f(3).

 

[Sneakatone]

Physics help on stopping distance.

With the brakes fully applied , a 1470 kg automobile decelerates at the rate of 7.5 m/s^2.
a) what is the magnitude of the breaking force acting on the automobile?
1470 *7.5=11025N ( I think this is correct)

b)if the initial speed is 90km/h (25 m/s) what is the stopping distance?
im thinking 25/7.5=3.33s
3.33*25=83.25 m

c)what is the work done by the breaking force at 90km/h.
part a) *part b)=J

d) what is the change in kinetic energy of the automobile?
1/2mv^2

I believe I know how to do the proceeding parts but I need the previous values.

 

[haruspex]

You really should start a new thread for a new question.

With the brakes fully applied , a 1470 kg automobile decelerates at the rate of 7.5 m/s^2.
a) what is the magnitude of the breaking force acting on the automobile?
1470 *7.5=11025N ( I think this is correct)

Yes. And the word is "braking".

b)if the initial speed is 90km/h (25 m/s) what is the stopping distance?
im thinking 25/7.5=3.33s
3.33*25=83.25 m

But it's not doing 25m/s for the whole 3.33s. Do you know an equation relating distance and time for constant acceleration? (There's a t² in it.)

c)what is the work done by the breaking force at 90km/h.
part a) *part b)=J

Yes. Note that it's simpler here because the force is constant.

d) what is the change in kinetic energy of the automobile?
1/2mv^2

Yes.

 

[Sneakatone]

I thought I did , I posted it here on accident. sorry.
d = vt + (1/2)at2
https://www.physicsforums.com/showthread.php?p=4300207#post4300207