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Standard X

Physics

Relation Between MA and VR

The work done...

Question

The work done in pulling up a block of wood weighing $2 k N$ for a length of $10 m$ on a smooth plane inclined at an angle of $15^{o}$ with the horizontal is:

5.17 k J

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6.42 k J

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7.03 k J

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9.82 k J

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Solution

The correct option is A $5.17 k J$
The component of displacement in vertical direction: $h c o s (90 - Ф) = 10 m e t e r s * C o s (90 - 15) d e g = 2.588 m e t e r s$
$W o r k d o n e = F o r c e * d i s p l a c e m e n t = 2 k N * 2.588 m e t e r s = 5.176 k J$

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The work done in pulling up a block of wood weighing 2kN for a length of 10m on a smooth plane inclined at an angle of 15o with the horizontal is:

The correct option is A 5.17kJThe component of displacement in vertical direction: hcos(90−Ф)=10meters∗Cos(90−15)deg=2.588metersWorkdone=Force∗displacement=2kN∗2.588meters=5.176kJ

The work done in pulling up a block of wood weighing $2 k N$ for a length of $10 m$ on a smooth plane inclined at an angle of $15^{o}$ with the horizontal is:

The correct option is A $5.17 k J$
The component of displacement in vertical direction: $h c o s (90 - Ф) = 10 m e t e r s * C o s (90 - 15) d e g = 2.588 m e t e r s$
$W o r k d o n e = F o r c e * d i s p l a c e m e n t = 2 k N * 2.588 m e t e r s = 5.176 k J$