Physics : force versus displacement?

AI Thread Summary
The discussion focuses on calculating kinetic energy and speed of a 2.0 kg object based on a force versus displacement graph. Users initially miscalculated work done, assuming the area under the graph was a rectangle, but later clarified that it involves a trapezium shape, requiring the addition of a triangle's area for accurate results. The correct approach involves calculating the total work done as the area under the curve up to each distance, with adjustments for negative forces affecting speed. The final calculations for kinetic energy and speed were confirmed, with users noting that negative work would reduce speed at larger displacements. The conversation emphasizes the importance of accurately interpreting the graph's area for determining work and subsequent kinetic energy.
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

1CQ8P.jpg

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.
 
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Plasm47 said:
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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

1CQ8P.jpg

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.

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.
 
PeterO said:
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.

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.
 
Plasm47 said:
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.

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?
 
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PeterO said:
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?

Please note I updated the last distance to 90 km - I had contemplated using just 1 hour.
 
Plasm47 said:
--------------------------------------------------------------------------------

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

1CQ8P.jpg

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.

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..
 
PeterO said:
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?

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.
 
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Plasm47 said:
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.

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.
 
PeterO said:
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.

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