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crimeas [40]
2 years ago
13

Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on th

e left weighs 60kg and is sitting three meters away from the fulcrum at the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. The mass of the board is evenly distributed so that its center of mass is over the fulcrum. What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground? 
a. 4m
b. 5m
c. 2m
d. 3m​
Physics
1 answer:
Lapatulllka [165]2 years ago
4 0

Answer:

Option A. 4 m

Explanation:

Please see attached photo for diagram.

In the attached photo, X is the distance from the centre to which the student on the right must sit in order to balance the seesaw.

Clockwise moment = X × 45

Anticlock wise moment = 3 × 60

Clockwise moment = Anticlock wise moment

X × 45 = 3 × 60

X × 45 = 180

Divide both side by 45

X = 180 / 45

X = 4 m

Thus, the student must sit at 4 m from the centre.

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To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

W = F * d

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The distance in this case is a composition between number of steps and the height. Then,

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On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

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W = 50*9.8*(0.3*30)

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PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

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Here,

V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow3 steps in one second

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a = -45*10^{-3}m/s^2

At this point we can calculate the time, which is,

t = \frac{\Delta V}{a}

t = \frac{0-0.9}{-45*10^{-3}}

t = 20s

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

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