JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 \(\left(g=10\, m s^{-2}\right)\)
- A \(200\)
- B \(150\)
- C \(275\)
- D \(325\)
Answer & Solution
Correct Answer
(B) \(150\)
Step-by-step Solution
Detailed explanation
\(W _{ F }=\frac{1}{2} mv ^{2}= mgh\) \(F ( S )= mgh\) \(F (0.2)=(0.15)(10)(20)\) \(F =150 N\)
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