Work Energy Power

JEE Main level, numerical answer, medium

110 of 183

Question 110 of 183

A cricket ball of mass 0.15kg0.15kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20m20m after leaving the machine. If the part pushing the ball applies a constant force FF on the ball applies a constant force FF on the ball and moves horizontally a distance of 0.2m0.2m while launching the ball, the value of F(inN)F(inN) is (g=10ms−2)\left(g=10m{s}^{-2}\right)

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Answer: 150

From work energy theorem F(0.2)−mg(20)=0F\left(0.2\right)-mg\left(20\right)=0
F=mg(20)0.2F=mg\dfrac{\left(20\right)}{0.2}
=mg(2002)=mg\left(\dfrac{200}{2}\right)
=0.15×10×2002=0.15\times 10\times \dfrac{200}{2}
=150.00N=150.00N

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