A particle of mass 100 g is projected vertic | vedclass

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Question

(B) The maximum height reached by the particle is given by $h = \frac{u^2}{2g}$.Substituting the values,$h = \frac{5^2}{2g} = \frac{25}{2g}$.The work

Vedclass · Accepted Answer

$-1.25$

Vedclass · Answer

(B) The maximum height reached by the particle is given by $h = \frac{u^2}{2g}$.Substituting the values,$h = \frac{5^2}{2g} = \frac{25}{2g}$.The work done by the force of gravity when the particle moves up is $W_{up} = -mgh = -m \cdot g \cdot \frac{25}{2g} = -1.25 \ J$.When the particle returns to the starting point,the displacement is zero.However,the question asks for the work done by gravity during the entire trip (up and down).Work done by gravity during the upward journey is $W_{up} = -mgh = -1.25 \ J$.Work done by gravity during the downward journey is $W_{down} = +mgh = +1.25 \ J$.The total work done by gravity for the complete round trip is $W_{total} = W_{up} + W_{down} = -1.25 + 1.25 = 0 \ J$.However,if the question implies the work done during the upward journey until it reaches the maximum height,the answer is $-1.25 \ J$.

$A$ particle of mass $100 \ g$ is projected vertically upwards with a speed of $5 \ m/s$. What is the work done by the force of gravity during the time the particle returns to the starting point? (in $J$)

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