A box weighing 2000 N is to be slowly slid through 20 m on a straigh track having friction coefficient 0.2 with the box. A find the work done by the person pulling the box wilth a chain at angle `theta` with the horizontal. B. find the work when the person has chosen a value of `theta` which ensures him the minimum magnitude of the force.

Answer

A, B, D

Explanation

Weight =2000 N,S=20 m `mu=2`

a. `R+psintheta-2000=0`………i
`Pcostheta-0.2R=0`………ii
From i and ii
`Pcostheta-0.2(2000-Psintehta)=0`
`P(costheta+0.2sintheta)=400 `
`P=400/(costheta+0.2sinthet)` ......iii
So, work done by the person,
W=PScostheta`
`=(8000costheta)/(costheta+0.2sintheta)`
`=8000/(1+0.32tantheta)`
`=40000/(5+tantheta)`......iv
b. From minium magnitude of force foerm equation i
`d/(dk)(costheta+0.2sintheta)=0`
`rarr tantheta=0.2`
putting the value in equation i
`W=40000/(5+tantheta)`
=40000/(5+0.2)~~7692J`