Finding velocities in F vs x graph

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SUMMARY

The discussion focuses on calculating the work done by a variable force on a 3.00 kg object as it moves along a position graph. The work for segments from x = 0 to x = 4.00 m, x = 5.00 m to x = 10.0 m, and x = 11.0 m to x = 15.0 m is determined to be 5J, 15J, and 5J respectively. The final speeds at x = 5.00 m and x = 15.0 m are calculated to be 2.29 m/s and 0.5 m/s, respectively. The discussion emphasizes the need for accurate force calculations and the correct application of the work-energy principle.

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Homework Statement


A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below.
p5-56.gif


Find the work done by the force on the object as it moves as follows:
(a) from x = 0 to x = 4.00 m
(b) from x = 5.00 m to x = 10.0 m
(c) from x = 11.0 m to x = 15.0 m.
(d) If the object has a speed of 0.500 m/s at x = 0, find:
speed at x = 5.00 m
speed at x = 15.0 m


The Attempt at a Solution


a. W=2.5N (4m) / 2 = 5J
b. W=3N (5m) = 15J
c. W=2.5N (4m) / 2 = 5J
d. W=1/2 mv2 - 1/2 mvo2

x=5m 15J = 3kg v - (3kg)(0.500m/s)
v=2.29 m/s

x=15m 0=3kg v - (3kg)(0.500m/s)
v=0.5 m/s
 
Last edited:
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mandy9008 said:

The Attempt at a Solution


a. W=2.5N (4m) / 2 = 5J
I would not use 2.5 N for the force. I suggest that you find an equation for the straight line in the form F = mx and calculate the force at 5 m.
b. W=3N (5m) = 15J
Looks good.
c. W=2.5N (4m) / 2 = 5J
Same comment as part (a)
d. W=1/2 mv2 - 1/2 mvo2

x=5m 15J = 3kg v - (3kg)(0.500m/s)
v=2.29 m/s

x=15m 0=3kg v - (3kg)(0.500m/s)
v=0.5 m/s
What happened to the 1/2 and the square of the speed in the expressions above?
 

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