Applying W=F*s for variable force

In summary: In practice, it is not always possible to get these quantities exactly, so using an approximation is usually acceptable.The average force is defined as the average of the forces over a certain time interval, and the average acceleration is the average of the accelerations over a certain time interval.It is not always possible to get these quantities exactly, so using an approximation is usually acceptable.
  • #1
UnknownGuy
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0

Homework Statement


A car of mass 2000kg moves along a horizontal road against a constant resistance of manitude (P)N. The total work done by the engine in increasing its speed from 4ms^-1 to 5.5ms^-1 while it moves a distance of 60m is 30000J. Find P.

Homework Equations


ΔEk+WP=WE

The Attempt at a Solution


Straightforward question. The correct solution is as follows:
ΔEk+WP=WE

1/2(2000)(5.5^2-4.0^2)+60P=30000
P=262.5J

However, another solution was proposed:

The work done by the engine is 30000J and the distance it moved was 60m so the average force it exerted was 30000/60=500N

500-P=ma
finding a using v^2=u^2+2as, and then multiplying it by 2000 (m) gives ma=237.5
P=262.5N

This alternative solution gives the same answer. However, a classmate pointed out that it is incorrect because he said it assumed a constant driving force, which is a wrong assumption. I think that dividing the total work done (30000J) by the total distance moved will give the AVERAGE value of this varying driving force and the value of a is the is the AVERAGE value of this varying acceleration. Is there a flaw in this method? If so, what is it?

Thanks in advance
 
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  • #2
UnknownGuy said:
it is incorrect because he said it assumed a constant driving force,
That is a valid criticism of the method, but it turns out not to matter in this case.
Indeed, v2=u2+2as is one of the SUVAT equations, and as a set those are only supposed to be for constant acceleration, i.e. the same assumption. However, that particular SUVAT equation is just energy conservation with mass canceled out, and so does not depend on constant acceleration.
The cleanest method would therefore be to use energy conservation rather than SUVAT.
 
  • #3
Your force balance equation is $$F-P=m\frac{dv}{dt}$$. If we multiply both sides of this equation by ##\frac{ds}{dt}=v##, we obtain:
$$F\frac{ds}{dt}-P\frac{ds}{dt}=mv\frac{dv}{dt}=\frac{m}{2}\frac{dv^2}{dt}$$If we next integrate this equation between 0 and t, we obtain:$$\int_0^s{Fds}-Ps=m\frac{v^2(t)-v^2(0)}{2}$$Dividing both sides of this equation by s yields:$$\frac{1}{s}\left[\int_0^s{Fds}\right]-P=m\frac{v^2(t)-v^2(0)}{2s}$$If we define the average force as $$\bar{F}=\frac{1}{s}\left[\int_0^s{Fds}\right]$$and the average acceleration as $$\bar{a}=\frac{v^2(t)-v^2(0)}{2s}$$we obtain:$$\bar{F}-P=m\bar{a}$$But this interpretation depends strictly on defining the average force and the average acceleration in this very specific way.
 

What is the formula for calculating work using variable force?

The formula for calculating work using variable force is W = F*s, where W represents work in joules (J), F represents force in newtons (N), and s represents displacement in meters (m).

How do you determine the direction of work when using variable force?

The direction of work when using variable force is determined by the direction of the force being applied. If the force is in the same direction as the displacement, then the work is positive. If the force is in the opposite direction of the displacement, then the work is negative.

Can you explain the concept of work using variable force with an example?

Yes, an example of work using variable force would be pushing a shopping cart. As you push the cart, you are applying a variable force to it. The displacement of the cart is in the same direction as the force being applied, so work is being done. However, as you stop pushing and the cart comes to a rest, the force being applied becomes zero and no work is being done.

What is the relationship between force and work when using variable force?

The relationship between force and work when using variable force is directly proportional. This means that as the force increases, the work done also increases, and vice versa. This relationship is represented by the formula W = F*s, where the force (F) and displacement (s) are directly proportional to the work (W) being done.

How is work calculated when using variable force in a system with multiple forces?

When using variable force in a system with multiple forces, the work is calculated by considering each force separately and then adding all of the individual work values together. This is because work is a scalar quantity and can be added algebraically. The final work value will depend on the net force acting on the system and the resulting displacement.

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