Question

A particle of mass$100g$ is thrown vertically upwards with a speed of $5$ m/s. The work done by the force of gravity during the time, the particle goes up is

• A. $-0.5J$
• B. $-1.25J$
• C. $1.25J$
• D. $0.5J$

Answer 1

The correct option is B

$-1.25J$

Step 1: Given data and assumptions.

Mass of the particle,

$\begin{array}{rcl}m& =& 100g\\ & =& 0.1kg\end{array}$

Initial speed, $v_i=5$m/s

Step 2: Finding the work done by the force of gravity when the particle goes up.

Since, we know that the work done by the force of gravity is the kinetic energy while moving up is,

$W_g={\left(KE\right)}_f-{\left(KE\right)}_i$

Where, $W_g$ is work done by force of gravity, ${\left(KE\right)}_f$ is final kinetic energy, ${\left(KE\right)}_i$ is initial kinetic energy.

$W_g=\frac{1m{v}_f^2}{2}-\frac{1m{v}_i^2}{2}$

Where $v_i$ is the initial velocity of the particle, $v_f$ is the final velocity of the particle

$\Rightarrow$$W_g=0-\frac{1\times 0.1\times 5^2}{2}$ $\left[\because v_f=0asgoesup\right]$

$\Rightarrow$$W_g=-1.25J$

Work done is negative indicating that the force and displacement are in opposite directions.

Thus, the work done by the force of gravity is ${\mathit{W}}_{\mathbf{g}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{}\mathit{J}$

Hence, option B is correct.