A-man-pushes-a-box-of-40kg-up-an-incline-of-15-if-the-man-applies-a-horizontal-force-200N-and-the-box-moves-up-the-plane-a-distance-of-20m-at-a-constant-velocity-and-the-coefficient-of-friction-is-0-

Question Number 34703 by NECx last updated on 10/May/18

A m a n p u s h e s a b o x o f 40 k g u p a n

i n c l i n e o f 15 °, i f t h e m a n a p p l i e s

a h o r i z o n t a l f o r c e 200 N a n d t h e

b o x m o v e s u p t h e p l a n e a d i s t a n c e

o f 20 m a t a c o n s t a n t v e l o c i t y a n d

t h e c o e f f i c i e n t o f f r i c t i o n i s 0.1,

f i n d

a) t h e w o r k d o n e b y t h e m a n o n t h e

b o x

b) w o r k d o n e a g a i n s t f r i c t i o n

Answered by tanmay.chaudhury50@gmail.com last updated on 10/May/18

w o r k d o n e b y t h e m a n i s

(m g s i n θ + μ m g c o s θ) \times s

= (40 \times 9.8 \times s i n 15^{o} + 0.1 \times 40 \times 9.8 \times c o s 15^{o}) \times 20

b) w o r k d o n e a g a i n s t f r i c t i o n

= μ m g c o s θ \times s