A Russian athlete lifts 300kg to a height of 2.5 m in 2 seconds. What is the work done? And what power is exerted?
1 answer:
We want to find the work done and power exerted, let’s start with work first.
We know that the equation for work is: W = F * D. We need to find the force which we can find by using: F = M * A.
Mass: 300kg
Acceleration (using equation from photo): 1.25 m/s^2
(The equation says x but can be used with y values)
If you are confused about how I found the acceleration; I plugged in 2.5 for the final y value, 0 for the initial y value, 0 for the initial velocity and 4 for t squared.
To solve, for acceleration it’s a matter of simple algebra. You can subtract the initial y position and the initial velocity from the final y position because they are 0. This leaves you with 2.5 m = 1/2a * t^2, from here I multiplied 2.5 by 2 to get rid of the 1/2. Now I have 5 = a * t^2. T^2 is just 2 squared, so four. Simply divide 5 by 4, and boom, you get 1.25 m/s^2.
Force = 300 kg * 1.25 m/s^2 = 375 Newtons
So, work = 500 N * 2.5 m = 1000 Joules
Power: W/t
So, Power = 1000 J / 2 seconds = 500 Watts
Hope this helps!
We know that the equation for work is: W = F * D. We need to find the force which we can find by using: F = M * A.
Mass: 300kg
Acceleration (using equation from photo): 1.25 m/s^2
(The equation says x but can be used with y values)
If you are confused about how I found the acceleration; I plugged in 2.5 for the final y value, 0 for the initial y value, 0 for the initial velocity and 4 for t squared.
To solve, for acceleration it’s a matter of simple algebra. You can subtract the initial y position and the initial velocity from the final y position because they are 0. This leaves you with 2.5 m = 1/2a * t^2, from here I multiplied 2.5 by 2 to get rid of the 1/2. Now I have 5 = a * t^2. T^2 is just 2 squared, so four. Simply divide 5 by 4, and boom, you get 1.25 m/s^2.
Force = 300 kg * 1.25 m/s^2 = 375 Newtons
So, work = 500 N * 2.5 m = 1000 Joules
Power: W/t
So, Power = 1000 J / 2 seconds = 500 Watts
Hope this helps!
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Answer:
Displacement: 6.71 m, Direction: 63.4 degrees north of east
Explanation:
In the attached image we can aprecciate each one of the movements of the parade. Let's say that the parade started from the origin (point (0,0)) then it moves to the east 4 blocks it means now the parade is located at point (4,0).
Then the parade went to the south three blocks, so it moves to the coordinate (4,-3). After this the parade went to the west one block so the new coordinate point is (3, -3).
And finally the movement of the 0 parade was 9 blocks to the north. It means the final point is now (0,9) - (3,-3) = (3,6)
And the displacement will be defined by the folliwing vector operation:
We know that the magnitude of the displacement vector is defined by the phytagoras theorem
And the angle will be defined by:
tan(beta)=3/6
beta = tan^-1(6/3)
beta = 63.43°
