Calculate Minimum Horsepower to Drag 340 kg Box at 1.45 m/s

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SUMMARY

The minimum horsepower required to drag a 340 kg box at a speed of 1.45 m/s on a level floor with a coefficient of friction of 0.50 is calculated using the formula Power = Work/time. The force of friction is determined to be 1667.7 N, which is essential for calculating the work done by the motor. Since the box moves at a constant velocity, the net force is zero, indicating that the motor must provide a force equal to the friction force to maintain motion. The kinetic energy of the box is 357.425 J, which aids in understanding the energy dynamics involved.

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Homework Statement



What minimum horsepower must a motor have to be able to drag a 340 kg box along a level floor at a speed of 1.45 m/s if the coefficient of friction is 0.50?

Homework Equations



Power = Change in Energy/time = Work/time
Change in energy = Final energy - Initial energy
Work = Net Force x displacement
Force of Friction = Normal force x coefficient of friction = mg x coefficient of friction
Kinetic energy = 1/2 x m x v^2
Change in gravitational potential energy = mgh

The Attempt at a Solution


I know that we should always start with the equation Power = Work/time or Power = Change in energy/time, and then sub in values. The problem is that of the equations given above, none of them seem to work. We are not given displacement. I found the friction force = 1667.7 N but I'm unsure as to how to find net force. Kinetic energy = 357.425 J, but is that the initial/final energy and how do I find the other one?
I'm just really confused about this question...please help! You don't have to tell me the whole solution, just a hint would be nice.
 
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Hint..there is no net force since the box is moving at constant velocity...so then what is the force being provided by the motor?
Hint 2: there is another formula for power equivalent to W/t...if W = Fd , then P = Fd/t...what is d/t?
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