Q: A block of weight W is dragged across the horizontal floor from A to B by the constant vertical force P acting at the end of the rope. Calculate the work done on the block by the force P = ( √3 + 1)N.Assume that block does not lift off the floor. (g = 10 m/s2 )
Ans: 4 J
Solution: W = P x
$W = P[\frac{h}{sin30^o} – \frac{h}{sin60^o} ] $
$W = P[\frac{h}{1/2} – \frac{h}{\sqrt{3}/2} ] $
$W = 2Ph[1 – \frac{1}{\sqrt{3}} ] $
$W = 2Ph[\frac{\sqrt{3}-1}{\sqrt{3}}] $
$W = 2(\sqrt{3}+1)\sqrt{3}[\frac{\sqrt{3}-1}{\sqrt{3}}] $
W = 4 J