A crane lifts a 425 kg steel beam vertically upward a distance of 66 m. How much work does the crane do on the beam if the beam
1 answer:
Answer:
W = 311074.5 [J]
Explanation:
In order to solve this problem we must analyze two parts, in the first part by means of Newton's second law we can determine the acceleration of the beam, remembering that the sum of the forces is equal to the product of mass by acceleration.
∑F = m*a
F = forces acting on the beam [N]
m = mass = 425 [kg]
a = acceleration = 1.8 [m/s²]
The forces acting on the beam are the force of the crane up (positive) and the weight of the beam down (negative)
Now in the second part, we use the definition of work, which is equal to the product of the force applied in the direction of displacement, that is, the product of force by distance.
where:
W = work [J]
F = force = 4713.25 [N]
d = distance = 66 [m]
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Answer:
+5.7 m/s
Explanation:
According to the law of conservation of momentum is that the momentum before the collision is equal to the momentum after the collision. In an equation form it would look like this:
M₁V₁+M₂V₂ = M₁V₁'+M₂V₂'
Where:
M₁ = mass of object 1 (kg)
V₁ = velocity of object 1 before the collision (m/s)
V₁' = Final velocity of object 1 after the collision (m/s)
M₂ = mass of object 2 (kg)
V₂ = velocity of object 2 before the collision (m/s)
V₂' = Final velocity of object 2 after the collision (m/s)
According to your problem you have the following given:
M₁ = 5 g = 0.005kg
V₁ = 3 m/s
V₁' = -5m/s (It bounced off so it is going the other direction)
M₂ = 6g = 0.006kg
V₂ = -1 m/s (It is coming from the opposite direction of the 3-ball)
V₂' = ?
So we plug in what we know and solve for what we don't know.