Question

A PERSON HOLDS A BUCKET OF WEIGHT 60N HE WALKS 7M ALONG HORIZONTAL AND THEN CLIMB A VERTICLE DISTANCE OF 5M THE WAORKDONE BY THE MAN WILL BE

Answer 1

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

Answer 2

Explanation:

hope this answer helps

so the total work done is 300J