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A stone of mass \( 0.05 \mathrm{~kg} \) is thrown vertically upwards. What is the direction and magnitude of
net force on the stone during its upward motion?
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net force on the stone during its upward motion?
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Verified Answer
The correct answer is:
\( 0.49 \mathrm{~N} \) vertically downwards
Given, mass $\mathrm{m}=0.05 \mathrm{~kg}$
When the stone is thrown upwards, force due to gravity acts on the stone vertically downwards, it is given as
$F=m g=0.05 \times 9.8=0.49$
Magnitude of net force $=0.49 \mathrm{~N}$
Direction of net force is vertically downwards.
When the stone is thrown upwards, force due to gravity acts on the stone vertically downwards, it is given as
$F=m g=0.05 \times 9.8=0.49$
Magnitude of net force $=0.49 \mathrm{~N}$
Direction of net force is vertically downwards.
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