Velocity from parabolic force time graph

[rasen58]

Find the speed of a 3 kg mass when it reaches the 10 m mark. The net force is in the same direction as the motion. Moving on straight line (rectilinear)

 

I think I'd have to use net work = change in kinetic energy theorem here.

Since the net work would be the area under the force position graph, I counted the boxes under the graph.

So I tried just counting all the boxes and got (around?) 39 boxes. I then multipled that by 20 since each box is 20 N and got 780 N

Since W = F *d = 780 N * 10 m = 7800 N*m

 

So 7800 = .5 m v^2

Solving for v, I got 22.8 m/s.

 

But the answer is actually 24.

I think that's due to me not counting the correct number of boxes? or is there a different way or a better way to approximate how many boxes are under the graph?

 

Thanks

[imatfaal]

0. It is a force distance graph not a force time graph

 

1. The area under the graph is not F - it is already Fd - ie work. You don't need to re-multiply it by 10metres - you have already done that by counting the boxes under the graph

2.

So 7800 = .5 m v^2

Solving for v, I got 22.8 m/s.

 

There is a maths error here.

 

[latex]7800 = \frac{1}{2}mv^2[/latex]

 

[latex]\sqrt{\frac{2*7800}{3}}=v=72.1 [/latex]

 

If however you use correct work per point 1 above

[latex]780 = \frac{1}{2}mv^2[/latex]

 

[latex]\sqrt{\frac{2*780}{3}}=v=22.80 [/latex]

 

3. There is a better way of approximating - though it is quite involved.

a. you know quadratic (you are told it is a parabola) is y=ax^2+bx+c

b. passes through origin so set x and y to zero and solve for c

c. assume that points (2,80) and (8,80) are on the line (they look perfect)

d. set simulatenous equations for y=ax^2+bx+c ; you have two points and already know c so these are solvable

e. you can check by solving your new equation and getting correct roots x=0 and x=10 and by differentiating at getting max at x=5

f. integrate your new equation from x=0 to x=10

g. plug new figure for area under the graph into Work Done = KE

h. I get just over 23 and a half.

i. I can go through process if I havent explained it properly

[rasen58]

Okay thank you for the corrections on all those points!