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 )?
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.
To find the average power delivered by the motor during the initial acceleration of the elevator, you need to consider both the time it takes and the distance traveled during that time. Here's the correct way to calculate it:
First, let's calculate the work done on the elevator during the initial acceleration. Remember that work is equal to force times distance:
W = Fd
The force F can be calculated using Newton's second law:
F = ma
where m is the mass of the elevator and a is the acceleration. In this case, the elevator starts from rest, so the acceleration during the initial part will be constant:
a = v / t
= 1.47 m/s / 3.13 s
≈ 0.47 m/s^2
Now, we can calculate the force:
F = ma
= 645 kg × 0.47 m/s^2
≈ 303.15 N
Next, let's calculate the work done:
W = Fd
= 303.15 N × 2.75 m
≈ 833.11 J
Finally, we can calculate the average power delivered by the motor:
P = W / t
= 833.11 J / 3.13 s
≈ 265.76 W
Therefore, the average power delivered by the motor during the initial acceleration of the elevator is approximately 265.76 Watts.
It seems that your initial calculation did not consider the distance traveled during the initial acceleration. By incorporating the distance in the calculation, you should obtain the correct answer.