How Much Power is Needed to Drag a Table Across a Brick Floor?

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To drag a 20.0 kg table across a brick floor at constant velocity, the force needed to overcome friction must be calculated using the coefficient of kinetic friction (0.600). The normal force is equal to the weight of the table, which is 20.0 kg multiplied by gravitational acceleration (9.8 m/s²), resulting in a frictional force of 117.6 N. The power required to drag the table over a distance of 14.0 m in 40.0 seconds is calculated using the formula P = F(d/t), yielding a power output of 68.6 W. It's crucial to focus on the frictional force rather than gravity when determining the work done. Understanding these calculations is key to solving the problem accurately.
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How much power must you exert to horizontally drag a 20.0 kg table 14.0 m across a brick floor in 40.0 s at constant velocity, assuming the coefficient of kinetic friction between the table and floor is 0.600?

Homework Equations



P=F(d/t), fk=mukN

The Attempt at a Solution



P=(20.0*9.8)(14.0/40.0) fk=(0.600*20.0*9.8)
=68.6W =117.6

117.6-68.6=49W



I really do not know how to do this problem, I just tried it in a way that made some sense to me, but it's obviously wrong. PLEASE HELP!
 
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When you drag something horizontally, you must do work to overcome the force of friction, not gravity.

your first equation P = F (d/t) tells you the power needed to push something with a force F, that's moving in the direction of that force with a constant speed v = d/t

The second equation F_k = \mu_k N tells you how big the friction is if you know the normal force N
 
So then do i need to find the work first and then whatever value I get from that use it to find the Power by using P=W/t?
 
You can do that, or calculate the speed first and then Power = force * speed.

The equation you gave P = F (d/t) does this all at once.

The important thing is to calculate the force you must exert first.
 
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