Question

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

Answer 1

Explanation:

Where, F =60N, S hori =7m , S vert =5 m

Let W1 and W2 be the work done in two cases.

Then W1=FS cos Θ =60*7*cos 90 =0

[Θ =90,because the force is vertical ]

W2 =60 *5*cos 0 =300 J

Total work done W =W1 +W2 =0+300 =300 J ANS