A body of mass 0.5 kg travels in a straight line with velocity \( v =ax^\dfrac{3}{2} \) where\( a = 5 m^\dfrac{-1}{2}s^{–1}\) What is the work done by the net force during its displacement from x = 0 to x = 2 m?

Asked by Abhisek | 1 year ago |  179

1 Answer

Solution :-

Mass of the body,m = 0.5 kg

Velocity of the body,

\( v= a x^\dfrac{3}{2}\)\(\)

here, a = \( 5 m^\dfrac{-1}{2}\) s-1.

Initial velocity at x = 0, v1 = a x 0 = 0

Final velocity at x = 2, v2 

 \(=a (2)^\dfrac{3}{2}=5\times\)  \( (2)^\dfrac{3}{2}\)

Work done by the system= increase in K.E of the body

\( \dfrac{1}{2}m(v_2^2-v_1^2) =\)  \( \dfrac{1}{2}\times 0.5 (5 \times (2)^\dfrac{3}{2})^2-0 \)\(\) 

= (\( \dfrac{1}{2}\)) x 0.5 x (25 x 8)=50 J

Answered by Abhisek | 1 year ago

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