A person holds a bucket of weight $60 N$ . He walks $7 m$ along the horizontal path and then climbs up a vertical distance of $5 m$ . The work done by the man is:

300 J

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420 J

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720 J

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Solution

The correct option is A $300 J$
Horizontal displacement give no work, because force acting normal to horizontal.

Force acting on vertical direction is $60 N$

Displacement in vertical direction is $5 m$

Work done, $W = F . d = 60 \times 5 = 300 J$

Hence, work done by man is $300 J$

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A person holds a bucket of weight 60 N. He walks 7 m along the horizontal path and then climbs up a vertical distance of 5 m. The work done by the man is:

The correct option is A 300 JHorizontal displacement give no work, because force acting normal to horizontal. Force acting on vertical direction is 60N Displacement in vertical direction is 5m Work done, W=F.d=60×5=300J Hence, work done by man is 300J

A person holds a bucket of weight $60 N$ . He walks $7 m$ along the horizontal path and then climbs up a vertical distance of $5 m$ . The work done by the man is:

The correct option is A $300 J$
Horizontal displacement give no work, because force acting normal to horizontal.

Force acting on vertical direction is $60 N$

Displacement in vertical direction is $5 m$

Work done, $W = F . d = 60 \times 5 = 300 J$

Hence, work done by man is $300 J$