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:
pf = 198.8 kg*m/s
θ = 46.8º N of E.
Explanation:
- Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.
- If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:
- We can do exactly the same for the initial momentum along the y-axis:
- The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:
- We can repeat the process for the y-axis, as follows:
- Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:
- In the same way, we can find the component of the final momentum along the y-axis, as follows:
- With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:
- The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:
- Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.
- We can find this angle applying the definition of tangent of an angle, as follows:
⇒ θ = tg⁻¹ (1.06) = 46.8º N of E