Work Done on Sled by Rope: 5.0s, 250N, 470N

AI Thread Summary
The sled is being pulled at a constant speed of 1.5 m/s by a rope with a tension of 250 N, inclined at 30 degrees. To calculate the work done by the rope over 5.0 seconds, only the horizontal component of the tension contributes, which is found using the formula 250 N * cos(30 degrees). The power delivered by the tension is then calculated by multiplying this horizontal component by the sled's speed. Finally, the work done is determined by multiplying the power by the time interval of 5.0 seconds. This approach effectively illustrates how to calculate work in a scenario involving inclined forces and horizontal motion.
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A sled is dragged along a horizontal path at a constant speed of 1.5 m/s by a rope that is inclined at an angle of 30 degrees with respect to the horizontal. The total weight of the sled is 470 N. The tension in the rope is 250 N. How much work is done by the rope on the sled in a time interval of 5.0 s?

http://www.webassign.net/grr/chapter-06/fig-024.gif
 
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evan4888 said:
A sled is dragged along a horizontal path at a constant speed of 1.5 m/s by a rope that is inclined at an angle of 30 degrees with respect to the horizontal. The total weight of the sled is 470 N. The tension in the rope is 250 N. How much work is done by the rope on the sled in a time interval of 5.0 s?

http://www.webassign.net/grr/chapter-06/fig-024.gif
HINTS:
a) Only the Tension horizontal component performs work since the motion is purely horizontal.
b) The Tension horizontal component is given by {(Tension)*cos(30 deg)}={(250 N)*cos(30 deg)}
c) {Tension Power Delivered} = {Tension Horizontal Component}*{Horizontal Speed}
d) {Tension Work Performed During Time Interval} = {Tension Power Delivered}*{Time Interval}
HOPE THIS HELPS.


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