A uniform brick is of size 6cm*8cm*8cm and has a mass of 2.5 kg. It is resting with its largest surface on the ground. What is the work done to bring it to rest on one of its smallest surface?
Answers
Given info : A uniform brick is of size 6cm × 8cm × 8cm and has mass 2.5 kg. It is resting with its largest surface on the ground.
To find : the work done to bring it to rest on one of its smallest surface.
Solution : when brick is in rest on its largest surface, height of brick = 6cm
so centre of mass = 6/2 = 3cm
work done at centre of mass, P₁ = mgh₁ = 2.5kg × 10m/s² × 0.03m
= 0.75 J
Now brick brings to rest on one of its smallest surface. Height of brick = 8cm
so, height of centre of mass = 8/2 = 4cm
Now work done at centre of mass, P₂ = mgh₂ = 2.5 kg × 10 m/s² × 0.04 m
= 1.00 J
Therefore net work done to bring it to rest on one of its smallest surface = P₂ - P₁
= 1 - 0.75 = 0.25 J
Answer:
Answer->0.245J
EXPLANATION:
>Initial height of brick =6cm= 0.06m
Height of center of mass of brick=0.03m
>Final height of brick=8cm= 0.08m
Height of center of mass of brick=0.04m
so, initial potential energy= 2.5×9.8×0.03
&,final potential energy=2.5×9.8×0.04
WORK DONE= final-initial
= 2.5×9.8×(0.04-0.03)
=0.245J
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