(B) Given,mass $m = 0.05 \,kg$.During the upward motion of the stone,the only force acting on it (ignoring air resistance) is the gravitational force.

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$0.49 \,N$ vertically downwards

(B) Given,mass $m = 0.05 \,kg$.During the upward motion of the stone,the only force acting on it (ignoring air resistance) is the gravitational force.The magnitude of the gravitational force is given by $F = m \times g$.Taking $g = 9.8 \,m/s^2$,we have:$F = 0.05 \,kg \times 9.8 \,m/s^2 = 0.49 \,N$.The direction of the gravitational force is always towards the center of the Earth,which is vertically downwards.Therefore,the net force is $0.49 \,N$ vertically downwards.

Class 11 Physics Newton's Laws of Motion and Friction The Common Forces and Equilibrium of Concurrent Forces

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$A$ stone of mass $0.05 \,kg$ is thrown vertically upwards. What is the direction and magnitude of the net force on the stone during its upward motion?

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A
$0.49 \,N$ vertically upwards
B
$0.49 \,N$ vertically downwards
C
$0.98 \,N$ vertically downwards
D
$9.8 \,N$ vertically downwards

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Newton's Laws of Motion and Friction

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The Common Forces and Equilibrium of Concurrent Forces

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$A$ stone of mass $0.05 \,kg$ is thrown vertically upwards. What is the direction and magnitude of the net force on the stone during its upward motion?

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