A Person is holding a bucket of water by applying a vertical force of 100N. He first moves a horizontal distance of 5 m and then climbs up through stairs a vertical distance of 10 m. calculate the total work done by the person on the bucket.

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Solution

$f o r h o r i z o n t a l m o t i o n, a n g l e b e t w e e n f o r c e a n d d i s p l a c e m e n t = 90 d e g r e e t h e r e f o r e W = f s \cos 90 = 0 f o r v e r t i c a l m o t i o n a n g l e = 180 d e g r e e W = 100 \times 10 \cos 180 W = - 1000 J o u l e n e t w o r k d o n e = - 1000 j o u l e$

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A Person is holding a bucket of water by applying a vertical force of 100N. He first moves a horizontal distance of 5 m and then climbs up through stairs a vertical distance of 10 m. calculate the total work done by the person on the bucket.

for horizontal motion, angle between force and displacement = 90 degreetherefore W = fscos90 = 0for vertical motion angle = 180 degreeW=100×10 cos180W = -1000 Joulenet work done = -1000 joule