A particle of mass 500 gm is moving in a straight line with velocity υ=bx52. The work done by the net force during its displacement from x = 0 to x = 4m is : ( Take b = 0.5m−32S−1).

ω=Δ K.E=12m[V12−V12]=12m[b×452]2−12m[0]2=12×0.5[(12)2×45]=24=16J=12(0.5)(14)×(45)=(12)(12)(14)(4)5=(142)=(43)=4×4×4=64J

A particle of mass 500 gm is moving in a straight line with velocity υ=bx52. The work done by the net force during its displacement from x = 0 to x = 4m is : ( Take b = 0.5m−32S−1).

ω=Δ K.E=12m[V12−V12]=12m[b×452]2−12m[0]2=12×0.5[(12)2×45]=24=16J=12(0.5)(14)×(45)=(12)(12)(14)(4)5=(142)=(43)=4×4×4=64J

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A particle of mass 500 gm is moving in a straight line with velocity $υ = b x^{\frac{5}{2}}$ . The work done by the net force during its displacement from x = 0 to x = 4m is : ( Take b = $0.5 m^{\frac{- 3}{2}} S^{- 1}$ ).

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$32 J$

$64 J$

$16 J$

$8 J$

answer is B.

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Detailed Solution

$\begin{array}{l} ω = Δ K . E = \frac{1}{2} m [V_{1}^{2} - V_{1}^{2}] \\ = \frac{1}{2} m {[b \times 4^{\frac{5}{2}}]}^{2} - \frac{1}{2} m {[0]}^{2} \\ = \frac{1}{2} \times 0.5 [{(\frac{1}{2})}^{2} \times 4^{5}] = 2^{4} = 16 J \\ = \frac{1}{2} (0.5) (\frac{1}{4}) \times (4^{5}) \\ = (\frac{1}{2}) (\frac{1}{2}) (\frac{1}{4}) {(4)}^{5} = (\frac{1}{4^{2}}) \\ = (4^{3}) = 4 \times 4 \times 4 = 64 J \end{array}$