A child applies a force F parallel to the x-axis to a 10 kg sled moving on the f
ID: 1261393 • Letter: A
Question
A child applies a force F parallel to the x-axis to a 10 kg sled moving on the frozen surface of a small pond. As the child controls the speed of the sled, the x-component of the force she applies varies with the x-coordinate of the sled as shown in the figure. Suppose the sled is initially at rest at x=0. You can ignore friction between the sled and the surface of the pond.
A.) Use the work-energy theorem to find the speed of the sled at 3.0 m.
B.) Use the work-energy theorem to find the speed of the sled at 9.0 m.
A child applies a force F parallel to the x-axis to a 10 kg sled moving on the frozen surface of a small pond. As the child controls the speed of the sled, the x-component of the force she applies varies with the x-coordinate of the sled as shown in the figure. Suppose the sled is initially at rest at x=0. You can ignore friction between the sled and the surface of the pond. A.) Use the work-energy theorem to find the speed of the sled at 3.0 m. B.) Use the work-energy theorem to find the speed of the sled at 9.0 m.Explanation / Answer
A) Using Work Eneergy theorem,
Work done = Change in K.E.
And we can find work done by this sketch . Area under the cuurve of F vs X gives work done for that region ( say
from x_initial to x_final)
from x = 0 to 3m
y at x = 3 : y = 10/8 x 3 = 3.75 m
work done = Area from 0 to 3 = 3 x 3.75 /2 = 5.625 J
work done = mv^2 /2 - 0
5.625 = 10v^2 /2 - 0
v = 1.06 m/s
b) similarly from x = 0 to x = 9 m
height at x = 9 : y = 10 - (10/4) x 1 = 7.5 m
area from 8 to 9 m = (10*4 /2) - (7.5*3 /2) = 8.75
Total Area from 0 to 9 = (10 x 8 /2) + (8,75) = 48.75
Work done = mv^2 /2 - 0
48.75 = 10v^2 /2
v = 3.12 m/s
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