Work Rate of Force on Block of 102 kg at 5.5 m/s

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SUMMARY

The discussion focuses on calculating the rate at which a 125 N force does work on a 102 kg block moving at a constant speed of 5.5 m/s across a horizontal surface. The key equation used is W = ΔE_k, where E_k is the kinetic energy defined as E_k = 1/2 mv². Despite the net work being zero due to constant speed, the question specifically inquires about the work done by the applied force, not the net force. The participant ultimately resolved their confusion regarding the distinction between these forces.

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Homework Statement


Across a horizontal floor,a ##102 kg## block is pulled at a constant speed of ##5.5\frac {m} {s}## by an applied force ##125N## directed ##38^°## above the horizontal.Calculate the rate at which the force does work on the block.

Homework Equations


##W=ΔE_k##
##E_k=\frac 1 2 mv^2##


The Attempt at a Solution


I didnt understand the question.Whats does the Calculate the rate at which the force does work on the block means.Net work must be zero cause object moves in constant speed.So there must be a fricition maybe ?

Thanks
 
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Arman777 said:

Homework Statement


Across a horizontal floor,a ##102 kg## block is pulled at a constant speed of ##5.5\frac {m} {s}## by an applied force ##125N## directed ##38^°## above the horizontal.Calculate the rate at which the force does work on the block.

Homework Equations


##W=ΔE_k##
##E_k=\frac 1 2 mv^2##


The Attempt at a Solution


I didn't understand the question.What does the Calculate the rate at which the force does work on the block means.Net work must be zero cause object moves in constant speed.So there must be a friction maybe ?

Thanks
The net work done is indeed zero, which can be seen because the speed is constant. As you say, there must be some other force.

But the question does not ask about the work done by the net force. It asks for the rate of work done by the 125 N force.
 
Thanks I solved
 

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