Respuesta :
Answer:
v = 2.21 m/s
Explanation:
The foreman had released the box from rest at a height of 0.25 m above the ground.
We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :
[tex]mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.25} \\\\v=2.21\ m/s[/tex]
So, its velocity at the bottom of the ramp is 2.21 m/s.
The crate's speed is 2.21 m/s.
Calculation of the speed:
Since the foreman had released the box from rest at a height of 0.25 m above the ground.
Here we need to determine the speed of the crate at the time when it reaches the bottom of the ramp.
Here we assume v be the velocity at the bottom of the ramp.
The formula of conversation of energy should be applied.
So, the speed should be
mgh = 0.5mv^2
v = √2gh
v = √2-9.8*0.25
v = 2.21 m/s
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