How Is Work Calculated with a Nonconstant Force?

[Rijad Hadzic]

Homework Statement

An object is subject to a nonconstant force f = (6x^3 -2x) ihat

such that the force is in Newtons when x is in meters. Determine the work done on the object as a result of this force as the object moves from x = 0 to x = 100 m

Homework Equations

w = f * d

The Attempt at a Solution

again not sure what I'm doing wrong here. Plugging in 100 m into f you get

5999800 m, and the total distance covered was 100 m, so multiplying the two you get 5.999 x 10^8 joules but book gives me 1.5 x 10^8 joules

I do not see how the books answer agrees with the equation w = f * d.

alternately I could use [itex] W = \Delta K + E_t [/itex] but I don't think using this equation is right in this case...

 

[berkeman]

Homework Statement

An object is subject to a nonconstant force f = (6x^3 -2x) ihat

such that the force is in Newtons when x is in meters. Determine the work done on the object as a result of this force as the object moves from x = 0 to x = 100 m

Homework Equations

w = f * d

The Attempt at a Solution

again not sure what I'm doing wrong here. Plugging in 100 m into f you get

5999800 m, and the total distance covered was 100 m, so multiplying the two you get 5.999 x 10^8 joules but book gives me 1.5 x 10^8 joules

I do not see how the books answer agrees with the equation w = f * d.

alternately I could use [itex] W = \Delta K + E_t [/itex] but I don't think using this equation is right in this case...

I'm pretty sure you need to use an integral for this problem. When the force changes as a function of distance, you need to integrate the force f(x) * dx over the range of distance x. Can you show us that work?

If you haven't use LaTeX before to show integral work, you can find a tutorial under INFO at the top of the page (click Help/How-To).

 

[Rijad Hadzic]

I'm pretty sure you need to use an integral for this problem. When the force changes as a function of distance, you need to integrate the force f(x) * dx over the range of distance x. Can you show us that work?

If you haven't use LaTeX before to show integral work, you can find a tutorial under INFO at the top of the page (click Help/How-To).

I see. So if acceleration is not constant, then we have to use integral method? and if the force IS constant, we can use equation W = F*d?

 

[Hilmy atha]

This involves basic calculus.. you need to find the Work equations first which is

6/4 × X^4 + X^2

now insert X = 100 and X = 0

1.5 10^8 +10^4 - (0)^4 -0^2Wait the answer is different from your book, why?._.
(Btw i can't give you the proper equations since I am on mobile, its too messy)

 

[berkeman]

This involves basic calculus.. you need to find the Work equations first which is

6/4 × X^4 + X^2

now insert X = 100 and X = 0

1.5 10^8 +10^4 - (0)^4 -0^2Wait the answer is different from your book, why?._.
(Btw i can't give you the proper equations since I am on mobile, its too messy)

I'm not sure this is helpful. Let's see how the OP's work looks first...

 

[berkeman]

I see. So if acceleration is not constant, then we have to use integral method? and if the force IS constant, we can use equation W = F*d?

You haven't been given any information about the acceleration. You are given f(x) and asked to find the work done over a range of x. That means you need to do the integral to figure out the total work over that distance.

 

[Rijad Hadzic]

I'm not sure this is helpful. Let's see how the OP's work looks first...

Hmm so I integrate function 6x^3 -2x

I get (6/4)x^4 - x^2
(6/4)(100)^4 - (100)^2 = 149990000

so that is my force. My distance is 100, so I multiply 149990000 by 100, and get 1.5 x 10^10

Still not my books answer :(. Why am I off by 10^10 - 10^8 :(?

 

[Rijad Hadzic]

You haven't been given any information about the acceleration. You are given f(x) and asked to find the work done over a range of x. That means you need to do the integral to figure out the total work over that distance.

Well its saying that the force is not constant, so I assumed acceleration to be non constant as well... does that seem right?

 

[Hilmy atha]

Hmm so I integrate function 6x^3 -2x

I get (6/4)x^4 - x^2
(6/4)(100)^4

so that is my force. My distance is 100, so I multiply149990000 by 100, and get 1.5 x 10^10

Still not my books answer :(. Why am I off by 10^10 - 10^8 :(?

No no, you got it right. Wow the answer was much closer to the book than i thoguht.

So,149990000 Joule IS your Work. Which rounds up to 1.5 × 10^8. The integral of Force IS work. Your equation has been converted to work equation

 

[Rijad Hadzic]

No no, you got it right. Wow the answer was much closer to the book than i thoguht.

So,149990000 Joule IS your Work. Which rounds up to 1.5 × 10^8. The integral of Force IS work. Your equation has been converted to work equation

Ahh okay finally I understand now.

Using the fact that integral of force is work, I didn't have to multiply by 100 at all.

Okay thank you guys. But also if you can answer my question:

"Well its saying that the force is not constant, so I assumed acceleration to be non constant as well... does that seem right?"

I appreciate everything!

 

[Hilmy atha]

Ahh okay finally I understand now.
Okay thank you guys. But also if you can answer my question:

"Well its saying that the force is not constant, so I assumed acceleration to be non constant as well... does that seem right?"

I appreciate everything!

Well yes.. if your Force changed, either the accel. or the mass changed. But you don't gain/lose mass along the way :p so yeah acceleration is the one that changes