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

0.98 N vertically downwards

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0.49 N vertically upwards

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9.8 N vertically downwards

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0.49 N vertically downwards

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Solution

The correct option is D
0.49 N vertically downwards

$F = m \times g = 0.05 \times 9.8 = 0.49 N, d o w n w a r d s$

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

The correct option is D 0.49 N vertically downwards F=m×g=0.05×9.8=0.49N,downwards

The correct option is D
0.49 N vertically downwards

$F = m \times g = 0.05 \times 9.8 = 0.49 N, d o w n w a r d s$