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Net Force and Net Work 
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#1
Feb2113, 07:45 PM

P: 143

1. The problem statement, all variables and given/known data
First, let us consider a sled of mass m being pulled by a constant, horizontal force of magnitude F along a rough, horizontal surface. The sled is speeding up. Let us now consider the situation quantitatively. Let the mass of the sled be m and the magnitude of the net force acting on the sled be F. The sled starts from rest. Consider an interval of time during which the sled covers a distance s and the speed of the sled increases from v_{1} to v_{2}. We will use this information to find the relationship between the work done by the net force (otherwise known as the net work) and the change in the kinetic energy of the sled Use W=Fscos(θ) to find the net work W_{net} done on the sled. Express your answer in terms of some or all of the variables m, v_{1} and v_{2}. 2. Relevant equations W=Fscos(θ) K=1/2mv^{2} ƩF=ma 3. The attempt at a solution The only thing I could do so far is the equation for F_{net}. The equation is unsimplified, so it is a little messy, but I attatched a picture of the F_{net} equation. Since W_{net}=F_{net}s, I assumed that if you multiply by s, the distance, then the s would cancel out and you would be left with the net force equation without the s in there. However, I'm very confused by what it is asking. It says that W=Fscos(θ), but the answer cannot have the variable θ in there. The only thing I could think of is that since it is being pulled on a flat surface, then cos(180°)=1, but since the sled is speeding up in the direction of the net force, work has to be positive. I cannot figure out what I should do next to try to solve this. Thank you in advance for the help. 


#2
Feb2113, 08:48 PM

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PF Gold
P: 4,759

You obviously would like s, the distance covered when the velocity reaches v, expressed as a function of v. Then you can say
Δs = s(v2)  s(v1) and then W = Fcos(θ)Δs. (it should be obvious what θ is here. It's the angle between the force vector and s. And, it's positive since F and s go in the same direction. You nearly got that one right). What is F in terms of a? Everyone knows that one! Newton knew it over 200 yrs. ago! So how do we get s(v)? Hint: use chain rule to get dv/ds and integrate both sides of an easy differential equation in s, v and constants. t is eliminated by the chain rule. Or maybe you already have a formula handy for coming up with s(v). When you get your answer you should compare it with the difference in kinetic energies of the sled at v1 and v2. 


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