A uniform beam of length 6m and mass 20kg is lying on horizontal floor. The work done in Joules to make it stand vertical on the floor is...
Answers
The work done in Joules to make it stand vertical on the floor is 588 Joules.
Explanation:
The length of the uniform beam, L = 6 m
The mass of the beam, m = 20 kg
Since the uniform beam is given to be lying on the horizontal floor, therefore, we can say that its initial potential energy i.e., initial P.E. = 0.
Now,
The final P.E. of the beam when it is standing vertical on the floor is given as,
= mg(L/2)sinθ
Here θ = 90° since the beam is standing vertical on the floor, g = 9.8 m/s² and substituting the other given values
= 20 * 9.8 * (6/2) * sin 90°
= 20 * 9.8 * 3 * 1
= 588 J
Thus,
The work done in Joules to make the uniform beam stand vertical on the floor is,
= [Final P.E.] – [Initial P.E.]
= 588 J – 0
= 588 J
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