Problem 6.38

6.38. a)

We first have to find the distance moved upwards in the first 3 s. Finding the acceleration of the elevator in that first three seconds yields

$a = \frac{v_{f}}{t} ~=~ .58 m/s^{2}$

then, the distance travelled in that time is given by

$\Delta y = \frac{1}{2} at^{2} ~=~ \frac{1}{2} (.58 m/s^{2})(3 s)^{2}$

$\Delta y = 2.625 m$

Now, to find the average power, we first find the work done on the system.

$W_{nc} = \Delta K + \Delta U = \frac{1}{2} mv_{f}^{2} - \frac{1}{2} mv_{i}^{2} + mg(\Delta y)$

$W_{nc} = 1.772 \times 10^{4} J$

and since the average power is given by

$P_{av} = \frac{W_{nc}}{t}$

we can find that

$P_{av} = \frac{1.772 \times 10^{4} J}{3 s} = 5910 W$

or P_av = 7.92 hp

b)

When the elevator is moving upward with a constant speed, the applied force is equal to the weight

F_app = mg = 6370 N

Therefore, since

P = Fv

we know that

P = (6370 N)(1.75 m/s) = 11100 W = 14.9 hp