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

Using work energy theorem,work done = change in Kinetic energy → $W_{total}=\Delta KE$
i.e. work done by gravity should be equal to change in the kinetic energy of the particle.
$$\begin{gathered} W_{gravity} = (K{E}_{final}-K{E}_{initial}) \\ {W}_{gravity} = (\frac{1}{2}m{v^2}_{final}-\frac{1}{2}m{v^2}_{initial}) \end{gathered}$$

given mass = 100g ; initial velocity = 5m/s; final velocity = 0; substitute in the above equation

$W_{gravity} = \left(0-\frac{{\displaystyle 1}}{{\displaystyle 2}}\times \left(\frac{{\displaystyle 100}}{{\displaystyle 1000}}\right){\left(5\right)}^2\right) =-1.25J$
Work done by force of gravity = -1.25J