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, v_{1} = a x 0 = 0

Final velocity at x = 2, v_{2}

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

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 agoA person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.

**(a)** How much work does she do against the gravitational force?

**(b)** Fat supplies 3.8 × 10^{7}J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

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