# Homework Help: Physics : force versus displacement?

1. Dec 3, 2011

### Plasm47

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In the graph below, forces labelled positive act in the direction of motion of the object, forces labelled negative oppose the motion. The object under consideration has a mass of 2.0kg and was initially at rest. Calculate its kinetic energy and speed when
a) d= 2.0m
b) d= 4.0m
c) d=6.0m
d) d= 8.0m

I know W is area under the graph, but not positive how to apply it. I assume the question is only asking for that moment in time, and NOT a time period.
so for (a) it would be W= 4.0N x 2.0m = 8J; which would also be Ek
and V= 2.83m/s
if this is correct then b and c follow the same method, so i got those right aswell.
Not sure of the procedure when the Force is negative.

2. Dec 3, 2011

### PeterO

Just one real problem. The area under the graph up to 2 m is not 4 x 2.

The area is not a rectangle - it is a trapezium 2 units wide, and 6 units high on one side and 4 units high on the other.
You can calculate it as the rectangle you used PLUS the triangle at the top if you like.

3. Dec 3, 2011

### Plasm47

I thought only the area under the variables (x=2, y=4) were considered in the calculations, thus meaning anything that exceeps 4N is irrelavent. Please explain furthur so I can understand the procedure.

4. Dec 3, 2011

### PeterO

If that were true, then if you take the first 6m, the area is zero, since it is 0N by then.

Look at the graph up to 6m - there is a large triangle under the graph up to that point. Base 6, height 6 so area 18units [18J actually]

If you consider only at the (2m,4N) position, there is no area. There is a line 4N high, but a line has zero thickness so the area of the line is zero.

The area under a velocity time graph gives the displacement. 2 hours after driving away from a traffic light [on a long straight road], my velocity is 45 km/h. does that mean I am 90 km from the light?

Last edited: Dec 3, 2011
5. Dec 3, 2011

### PeterO

Please note I updated the last distance to 90 km - I had contemplated using just 1 hour.

6. Dec 3, 2011

### PeterO

re sentence in red
Not really.
They do want to know the speed at that instant of time, but that is determined by what happens during the preceding time interval, so the area under the graph means the area under the graph up to that point..

7. Dec 3, 2011

### Plasm47

ok that makes sense. So that being said, i'll post what i got for (a)
a)to calculate work, solve for area under curve
W= l x w + (bXh)/2
W= 2m x 4N + (2m x (6N-4N)/2
W= 8J + 2J
w= 10 J
The Ek is equivalent to w since there is no work lost in the process.
Ek= 10J

now, solve for velocity
Ek= 0.5 (m)v2
v2= 2x10J/2kg
v2= 10
v= √10
v=3.16m/s (to two sig digs)

if this is correct, then (b), (c), and (d) will follow the same process. Also, i estimate the W will be larger for b and more so for c.

Last edited: Dec 3, 2011
8. Dec 3, 2011

### PeterO

That's better. And when you get to 8m, the "negative work" - the force will be slowing down rather than speeding up the mass - the speed will be knocked back to a familiar answer.

9. Dec 3, 2011

### Plasm47

Oh yeah! the kinetic energy and velocity in (d) are the same as those calculated in (b)
thanks for all your help, greatly appreciated!