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 ?

Question

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 ? (A) 0.49 N vertically upwards (B) 0.49 N vertically downwards (C) 0.98 N vertically downwards (D) 9.8 N vertically downwards

Tardigrade Admin · Accepted Answer

Given, m = 0.05 kg We know that, net force acting on a stone, fg=mg = 0.05 × (-9.8)=-0.40 N = 0.49 N vertically downwards

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

0.49 N vertically upwards

0.49 N vertically downwards

0.98 N vertically downwards

9.8 N vertically downwards

Given, $m = 0.05 k g$
We know that, net force acting on a stone,
$f_{g} = m g$
$= 0.05 \times (- 9.8) = - 0.40 N$
$= 0.49 N$ vertically downwards

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