Thursday
March 23, 2017

Post a New Question

Posted by on Monday, June 6, 2011 at 8:08pm.

A 645-kg elevator starts from rest and reaches a cruising speed of 1.47 m/s after 3.13 seconds. It moved 2.75 m during that time.

What average power (W) is delivered by the motor during the initial acceleration of the elevator during the first 3.13 seconds?

The correct solution is 5780 W but I am no where near this number. The equation I am using is P = (mv^2)/2t + mgv/2

(645 kg x 1.47 m/s)/2(3.13) = 151.46
(645 kg x 9.8 x 1.47)/2 = 4645.93
4645.93 + 151.46 = 4797 W

Should I be incorporating the distance traveled that is given (2.75 m )?

  • Physics - , Monday, June 6, 2011 at 8:27pm

    energy used= Force*distance+final KE
    = m(g+a)*2.75+1/2 m*1.47^2
    =645(9.8+1.47/3.13)*2.75 +1/2*645*1.47^2

    That should give you the energy used, then divide it by 3.13 to get watts.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question