JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 5)
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
150
Explanation
Initial velocity, v = $$\sqrt {2gh} $$
= $$\sqrt {2 \times 10 \times 20} $$
= 20 m/s
Now work done by the machine,
WF = $$\Delta $$k
$$ \Rightarrow $$ F.d = $$\Delta $$k
$$ \Rightarrow $$ F = $${{\Delta k} \over d}$$
= $${{{1 \over 2} \times 0.15 \times 400 - 0} \over {0.2}}$$
= 150 N
= $$\sqrt {2 \times 10 \times 20} $$
= 20 m/s
Now work done by the machine,
WF = $$\Delta $$k
$$ \Rightarrow $$ F.d = $$\Delta $$k
$$ \Rightarrow $$ F = $${{\Delta k} \over d}$$
= $${{{1 \over 2} \times 0.15 \times 400 - 0} \over {0.2}}$$
= 150 N
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