A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 m after leaving the machine. If the part pushing the ball applies a constant force F on the ball and moves horizontally a distance of 0.2 m while launching the ball, the value of F (in N) is (g = 10 ms–2) ____.
Answer (integer)
150
Solution
Initial velocity, v = $\sqrt {2gh}$
<br><br>= $\sqrt {2 \times 10 \times 20}$
<br><br>= 20 m/s
<br><br>Now work done by the machine,
<br><br>W<sub>F</sub> = $\Delta$k
<br><br>$\Rightarrow$ F.d = $\Delta$k
<br><br>$\Rightarrow$ F = ${{\Delta k} \over d}$
<br><br>= ${{{1 \over 2} \times 0.15 \times 400 - 0} \over {0.2}}$
<br><br>= 150 N
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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