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

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AI Thread Summary
The discussion revolves around calculating the rate at which a 125 N force does work on a 102 kg block being pulled at a constant speed of 5.5 m/s. Participants clarify that since the block moves at constant speed, the net work done is zero, indicating the presence of friction or another opposing force. However, the main focus is on the work done specifically by the 125 N force, not the net force. The original poster expresses confusion about the question but ultimately resolves their misunderstanding. The conversation highlights the importance of distinguishing between net work and the work done by individual 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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