The correct option is (B) : 60 J
vi = 3(0)2 + 4 = 4 ≅ x = 0
vF = 3(2)2 + 4 = 16 ≅ x = 2
\(W=ΔK=\frac{1}{2}m(16^2-4^2)\)
\(=\frac{1}{2}\times\frac{1}{2}(256-16)=\frac{240}{4}\)
\(=60\ J\)
A block of certain mass is placed on a rough floor. The coefficients of static and kinetic friction between the block and the floor are 0.4 and 0.25 respectively. A constant horizontal force \( F = 20 \, \text{N} \) acts on it so that the velocity of the block varies with time according to the following graph. The mass of the block is nearly (Take \( g = 10 \, \text{m/s}^2 \)): 
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