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andreyandreev [35.5K]
3 years ago
11

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

accelerates upward at 1.8 m/s2? Neglect frictional forces.
Physics
1 answer:
iogann1982 [59]3 years ago
5 0

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)

F_{crane}-(425*9.81)= 425*1.8\\F_{crane}=4713.25 [N]

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.

W=F*d

where:

W = work [J]

F = force = 4713.25 [N]

d = distance = 66 [m]

W=4713.25*66\\W=311074.5[J]

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Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

\omega_f = \omega_i + \alpha t

where

\omega_i = 0.300 rev/s is the initial angular velocity

\alpha = 0.895 rev/s^2 is the angular acceleration

Substituting t = 0.200 s, we find

\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s

Let's now convert it into rad/s:

\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

r=\frac{0.800}{2}=0.400 m

And so now we can find the tangential speed at t = 0.200 s:

v=\omega_f r =(3.01)(0.400)=1.2 m/s

2) 2.25 m/s^2

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

a_t = \alpha r

where \alpha is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2

A point on the tip of the blade has a distance of

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From the centre; so, the tangential acceleration is

a_t = (5.62)(0.400)=2.25 m/s^2

3) 3.6 m/s^2

The centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

a=\frac{1.2^2}{0.400}=3.6 m/s^2

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Answer:

Explanation:

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On a weather map, lines joining places with the same air pressure are called ___?
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Answer:

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Answer:

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Explanation:

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All centers will be closed on Friday 9/3 and Monday 9/6

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I HOPE IT'S HELP.......

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