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Step-by-Step Solution
Step 1: Identify the Physical Scenario
A particle is thrown vertically upward. We want to determine the work done by the gravitational force on the particle while it rises.
Step 2: Understand the Relevant Concept
The work done by gravity is given by
$$W = F \cdot d \cos\theta,$$
where $F$ is the force, $d$ is the displacement, and $\theta$ is the angle between the force and displacement. In a vertical upward throw, the force of gravity acts downward ($mg$) and the displacement is upward, making the work done by gravity negative.
Step 3: Express Work Done by Gravity in Terms of Energy
From energy conservation, the change in kinetic energy is equal to the net work done. When thrown vertically upward, the loss in kinetic energy equals the gain in potential energy. According to the work-energy theorem, the work done by the weight of the particle (gravity) is equal to the negative of the change in kinetic energy.
Initially, the particle has kinetic energy:
$$K.E = \frac{1}{2} m v^2.$$
The work done by gravity during the ascent is then:
$$W = - \Delta K.E = - \frac{1}{2} m v^2.$$
Step 4: Substitute the Given Values
Mass $m = 100 \,\text{g} = 0.1\,\text{kg}$.
Velocity $v = 5 \,\text{m/s}$.
Hence, the kinetic energy at the initial instant is:
$$K.E = \frac{1}{2} \times 0.1 \,\text{kg} \times (5 \,\text{m/s})^2 = 1.25 \,\text{J}.$$
The work done by gravity during the upward travel therefore becomes:
$$W = -1.25 \,\text{J}.$$
Step 5: Conclude the Result
The work done by the gravitational force on the particle while it goes up is -1.25 J, which matches the provided correct answer.
