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

  • Thread starter Arman777
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In summary, the question is asking for the rate at which the 125 N force does work on a 102 kg block moving at a constant speed of 5.5 m/s across a horizontal floor. Despite the net work being zero due to the constant speed, the question specifically asks for the rate of work done by the applied force.
  • #1
Arman777
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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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  • #2
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.
 
  • #3
Thanks I solved
 

1. What is the formula for calculating work rate of force?

The formula for calculating work rate of force is work rate = force x velocity. In this case, the force is the weight of the block, which can be calculated by multiplying the mass (102 kg) by the acceleration due to gravity (9.8 m/s^2). So, the formula for this specific situation would be work rate = (102 kg x 9.8 m/s^2) x 5.5 m/s = 5,643.6 Joules/second.

2. How is work rate of force related to the block's speed?

The work rate of force is directly proportional to the block's speed. This means that as the speed increases, the work rate of force also increases. In other words, the faster the block is moving, the more energy it is using per unit of time.

3. What are the units for work rate of force?

The units for work rate of force are Joules/second, also known as watts. This is a unit of power, which represents the rate at which work is done or energy is used.

4. Is work rate of force the same as power?

Yes, work rate of force and power are two terms used interchangeably to describe the same concept. Both refer to the rate at which work is done or energy is used.

5. How can work rate of force be increased for the block?

In order to increase the work rate of force for the block, either the force or the speed of the block must be increased. This can be done by either applying a larger force on the block or increasing its velocity. For example, if the force on the block is doubled, the work rate of force will also double. Similarly, if the speed of the block is doubled, the work rate of force will also double.

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