Answered

A block of ice weighs 500 N. What is the minimum work is required to push it up a 6 m long incline that rises 3m?

Answer :

The work done in lifting the weight on an inclined plane is 2598N

Explanation:

Given:

Weight, W = 500N

Distance, s = 6m

[tex]sin \alpha = \frac{Perpendicular}{hypotenuse}[/tex]

Where,

Perpendicular = 3m

Hypotenuse = 6m

Substituting the value:

[tex]sin \alpha = \frac{3}{6} \\\\sin \alpha = \frac{1}{2} \\\\\alpha = 30^o[/tex]

Work done = F X scosα

[tex]W = 500 X 6 X cos(30)\\\\W = 3000 X \frac{\sqrt{3} }{2} \\\\\\W = 1500\sqrt{3} \\\\W = 2598N[/tex]

Therefore, work done in lifting the weight on an inclined plane is 2598N

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