KCET · Physics · Communication System
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?
- A \( 0.49 \mathrm{~N} \) vertically upwards
- B \( 0.49 \mathrm{~N} \) vertically downwards
- C \( 0.98 \mathrm{~N} \) vertically downwards
- D \( 9.8 \mathrm{~N} \) vertically downwards
Answer & Solution
Correct Answer
(B) \( 0.49 \mathrm{~N} \) vertically downwards
Step-by-step Solution
Detailed explanation
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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