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

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SUMMARY

The work done by the rope on the sled is calculated using the horizontal component of the tension in the rope. Given a tension of 250 N at an angle of 30 degrees, the horizontal component is determined using the formula {(Tension)*cos(30 deg)}. The sled moves at a constant speed of 1.5 m/s for a time interval of 5.0 seconds, leading to the calculation of work done as {Tension Power Delivered} = {Tension Horizontal Component}*{Horizontal Speed} and subsequently {Tension Work Performed During Time Interval} = {Tension Power Delivered}*{Time Interval}. The final work done is 1,250 J.

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evan4888
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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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