Question

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 33.5° above the horizontal. The car accelerates uniformly to a speed of 2.15 m/s in 13.0 s and then continues at constant speed. (a) What power must the winch motor provide when the car is moving at constant speed? kW (b) What maximum power must the motor provide? kW (c) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of length 1,250 m? J Need Help? Read It

Answer 1

Answer:

a). P=11.04kW

b). Pmax=11.38 kW

c). Wt=6423.166kJ

Explanation:

The power of the motor when the speed is constant is the work in a determinate time.

$P=\frac{W}{t}$

The work is the force the is applicated in a distance so

W=F*d

replacing:

$P=F*\frac{d}{t}$ and $\frac{d}{t}$ determinate distance in time is velocity so

a).

$P=F*v$

$F=m*a\\F=m*g*sin(33.5)$

$P=950kg*9.8\frac{m}{s^{2}}*sin(33.5)*2.15\frac{m}{s}\\P=11047.846 W\\P=11.0478 kW$

b).

The maximum power must the motor provide, is the maximum force with the maximum speed of the motor in this case

The first step is find the acceleration so

$vi=0\frac{m}{s} \\vf=2.15 \frac{m}{s}\\vf=vi+a*t\\vf-vi=a*t\\ a=\frac{vf-vi}{t}= a=\frac{2.15\frac{m}{s}-0\frac{m}{s}}{13s}\\a=0.1653 \frac{m}{s^{2}}$

The maximum force is when the car is accelerating so

$Ft=Fa+Fg\\Ft=m*a+m*g*sin(33.5)\\Ft=950kg*0.1653\frac{m}{s^{2}}+950*9.8\frac{m}{s^{2}}*sin(33.5)\\Ft=5295.565 N$

so the maximum force is the maximum force by the maximum speed

$Pmax=Ft*v\\Pmax=5295.565N*2.15\frac{m}{s}\\Pmax=11385.46\\Pmax=11.3854kW$

c).

The total energy transfer without any friction is the weight move in the high axis y in this case, so is easy to know that distance

W=m*g*h

h=Length*sin(33.5)

W=m*g*Length*sin(33.5)

W=950 kg*9.8* 1250m*sin(33.5)

W=6423166.667 kJ

W=6423.166 kJ