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Work Power and Energy question

2006 · Shift 0 · Q127

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Work Power and Energy
2006 · Shift 0 · Q127

Work Power and Energy question

2006 · Shift 0 · Q127

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1

A particle of mass

100g

is thrown vertically upwards with a speed of

5m/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
$0.5J$

View written solutionFree

Correct answer: B

Given data

Mass of particle: $100\text{ g} = 0.1\text{ kg}$
Initial upward speed: $u = 5\text{ m/s}$

We need the work done by gravity during the upward motion.

Use work-energy theorem for gravity

As the particle goes upward, its speed decreases from $5\text{ m/s}$ to $0$ at the highest point.

The work done by gravity equals the change in kinetic energy:

$W_g = K_f - K_i$

Initial kinetic energy:

$K_i = \frac{1}{2}mu^2 = \frac{1}{2}(0.1)(5^2)$

$K_i = 0.05 \times 25 = 1.25\text{ J}$

Final kinetic energy at the top:

$K_f = 0$

Therefore,

$W_g = 0 - 1.25 = -1.25\text{ J}$

Interpretation

Gravity acts downward while displacement is upward, so the work done by gravity must be negative. This matches the result.

Option check

A: $-0.5\text{ J}$ ❌
B: $-1.25\text{ J}$ ✅
C: $1.25\text{ J}$ ❌
D: $0.5\text{ J}$ ❌

So the correct option is B.

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