Calculating Power Needed to Push a Box Across a Floor with Friction

[lailanni]

I've been trying this for an hour, please have mercy and help!

You push a 67 kg box across a floor where the coef. Kfriction is .55. The force you exert is horizontal. How much power is needed to push the box at a speed of .5m/s?

I've tried w=Ki-Wnc, but I guess that isn't it.

Thank you!

 

[xman]

To solve this problem sum the forces in the x and y directions. You should get
F_{x} = \frac{dp}{dt} + f_{s} and F_{y} = N= m g...(1)
calling this eq (1). So we may write from eq (1)
f_{s} = \mu_{s} m g ...(2)
Now we wish to know the power, which we may relate to the force exerted by the equation
P= F v ...(3)
So substitute eq (2) and eq (1) into eq (3), with the numerical values given in the problem. Note that the quantity
\frac{dp}{dt}
is the change in momentum with respect to time, if the force is a constant force (which I am assuming it is) this term is zero, which yields the result for eq (3) is
P = \mu_{s} m g v
which are all numerical values in problem statement. I hope this helps.
sincerely, x

 

[Hootenanny]

Work done = force x distance.
The force in this case is $F = \mu R$ where $\mu$ is the coefficent of friction.
Let the distance moved be 0.5m.
Power is $\frac{workdone}{time}$, the speed is 0.5m/s, therefore the time is 1 and in this case, power = workdone.