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2 years ago

A 570 kg elevator accelerates downwards at 1.5 m/s2 for the first 13 m of its motion.

Physics

1 answer:

jeka942 years ago

6 0

Mass of the elevator (m) = 570 Kg
Acceleration = 1.5 m/s^2
Distance (s) = 13 m
Let the force be F.
We know, F = ma,
Therefore, F = (570 × 1.5) N = 855 N
Angle between distance and force (θ) = 0°
We know, work done = F s Cos θ
Therefore, work done by the cable during this part
= (855 × 13 × Cos 0°) J
= (855 × 13 × 1) J
= 11115 J

Answer:

11115 J

Hope you could get an idea from here.

Doubt clarification - use comment section.

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The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

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The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

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