Given Force is 4xDisplacement, find work done

[Govind_Balaji]

Homework Statement

A force $F=4x$ is applied on an object. What is work done to move it from $x_1=2m$ to $x_2=4m$?

A)12 J
B)24 J
C)32 J
D)48 J

Homework Equations

The Attempt at a Solution

The displacement, $x=4-2=2m$
Force, $F=4x=4 \times 2=8N$
Work $W=\vec F \cdot \vec x=8 \times 2 = 16 J$
But it is not listed in options. Did I miss anything? I guess it because Force was given as 4 times x, which will result in meters than Newtons.

 

[ehild]

The force is not constant, so you have to calculate the work as $W=\int_2^4F(x)dx$

F=4x means that 4 has the proper unit, N/m.

 

[Govind_Balaji]

The force is not constant, so you have to calculate the work as $W=\int_2^4F(x)dx$

F=4x means that 4 has the proper unit, N/m.

Horrible! Calculus is not in the portion. anyway I know some basics of it and figured out the answer as 32-8=24 J. Thank you very much.

 

[ehild]

If you haven't studied Calculus yet, plot the function F(x) and find the area under the plot between the limits x=2 and x=4. It is a trapezoid now.

 

[Govind_Balaji]

I have one doubt. What is the correct definition of work?
My books say $W=\vec F \cdot \vec x$
But $W=\int F(x) dx$ seems more reasonable.

Also what if force is given as a function of time.

 

[ehild]

$W= \vec F \cdot \vec x$ if the force is constant and $\vec x$ is the displacement.
If the force changes with the position x, the elementary work done during a small displacement $\vec {dx}$ is $dW=\vec F \cdot \vec {dx}$, the scalar product of the force with the small displacement.More accurately
To get the whole work you have to integrate for the whole path..

If the force depends on time, you change for the variable t:
$dW=\vec F \cdot \vec {dx}= \vec F \cdot (\vec {dx}/dt )dt=\vec F \cdot \vec {v} dt=$

 

[Govind_Balaji]

$W= \vec F \cdot \vec x$ if the force is constant and $\vec x$ is the displacement.
If the force changes with the position x, the elementary work done during a small displacement $\vec {dx}$ is $dW=\vec F \cdot \vec {dx}$, the scalar product of the force with the small displacement.More accurately
To get the whole work you have to integrate for the whole path..

If the force depends on time, you change for the variable t:
$dW=\vec F \cdot \vec {dx}= \vec F \cdot (\vec {dx}/dt )dt=\vec F \cdot \vec {v} dt=$

Amazing!, how to work on the last equation? Is this correct?
$$dW=\vec F \cdot \vec v dt$$
$$\frac{dW}{dt}=\vec F \cdot \vec v$$[Can I just divide the equation by $dt$ like this one?\
$$\text{Power}=\vec F \cdot \vec v$$
Is this also right?
$$\begin{align*}\\\frac{dW}{dt}&=\vec F \cdot \vec v\\\\<br /> \frac{dW}{dt}&=\frac{m\cdot \vec{dv} }{dt}\cdot \vec v\\\\<br /> dW&=m\cdot dv^2<br /> \end{align*}\\$$

 

[Ravi4213]

Actually you have to take average of displacement means x=X1+X2/2 than you find the answer it will be 24.

 

[ehild]

Amazing!, how to work on the last equation? Is this correct?
$$dW=\vec F \cdot \vec v dt$$
$$\frac{dW}{dt}=\vec F \cdot \vec v$$[Can I just divide the equation by $dt$ like this one?\
$$\text{Power}=\vec F \cdot \vec v$$
Is this also right?

Yes, it is right. The power is the scalar product of force with velocity.. It is a very useful equation!

$$\begin{align*}\\\frac{dW}{dt}&=\vec F \cdot \vec v\\\\<br /> \frac{dW}{dt}&=\frac{m\cdot \vec{dv} }{dt}\cdot \vec v\\\\<br /> dW&=m\cdot dv^2<br /> \end{align*}\\$$

It is not right, as
$\vec v \cdot \frac{d \vec v}{dt} = 0.5 \frac {d(\vec v)^2} {dt}$,

So $dW=0.5 m d(\vec v)^2= d KE$

This is the Work-Energy Theorem. The change of kinetic energy is equal to the work done by the force (resultant of all external forces) . .

 

[Govind_Balaji]

Actually you have to take average of displacement means x=X1+X2/2 than you find the answer it will be 24.

I think x1, x2 are position rather than displacement.
Displacement is the change in position, isn't it?

 

[ehild]

I think x1, x2 are position rather than displacement.
Displacement is the change in position, isn't it?

x1 and x2 are positions. The displacement is x2-x1.