Question

A particle of mass 100g is thrown vertically upwards with a speed of $5\;m{s^{ - 1}}$. What is the work done by the force of gravity during the time the particle goes up?
A. $- 0.5\;J$
B. $- 1.25\;J$
C. $1.25\;J$
D. $0.5\;J$

Answer 1

When a particle of a certain mass is thrown vertically upward with a certain speed. Then the kinetic energy at projection point is converted as the potential energy of the particle while rising. Potential energy is based on mass, gravity and height.

Formula used:
Work done on the mass formula:
$w = m.g.h$
Where ${w}$ is the work done against the force
$m$ is the mass of the object
$h$ is the height (or) position
$g$ is the acceleration due to gravity.

Complete step by step solution:
Mass of the object $\left( m \right) = 100\;g$
Vertical speed of the object $\left( u \right) = 5\;m{s^{ - 1}}$
When the object is raised from the ground, the work is done against the gravitational force, the work done on the mass is given by
${w_g} = m.g.h\;..........\left( 1 \right)$
Where,
${w_g}$ is the work done against the force
$m$ is the mass of the object
$h$ is the height (or) position
$g$ is the acceleration due to gravity.

Substitute the known value in (1)
${w_g} = \left( {100 \times {{10}^{ - 3}}} \right) \times 9.81 \times h\;.........\left( 2 \right)$
$\therefore$Acceleration due to gravity, $g = 9.81$ m/$s^2$
The vertical height to which the particle gets rise is $h$
So,
$h = \dfrac{{{u^2}}}{{2g}}$
Where
$u$ is the initial velocity
$g$ is the acceleration due to gravity.
By substituting the known values
$h = \dfrac{{{{\left( 5 \right)}^2}}}{{2 \times 9.8}} = 1.27\;m \\\ h = 1.27\;m \\\$
Substitute the value of height $h$ in equation (2)
${w_g} = \left( {100 \times {{10}^{ - 3}}} \right) \times 9.81 \times 1.27 \\\ {w_g} = 1.25\;J \\\$
Since the work ${w_g}$ is done against the gravitation force.it is negative
Then,
$\- {w_g} = 1.25\;J \\\ {w_g} = - 1.25\;J \\\$
$\therefore$ The work done by the force of gravity during the time the particle goes up is $- 1.25\;J$. Hence, the options (B) is correct.

Note:
When in the absence of air resistance all the objects which are in free fall will hit the ground at the same time regardless of the mass of the substance. Actually, the acceleration due to gravity is always negative, that is $- 9.81\;m{s^{ - 2}}$. But when the symbol $g$ is used in any of the equations, the direction will be assumed and the absolute value will be used for calculations as $\left( { + 9.81\;m{s^2}} \right)$.