Kinetic energy from rest under uniform acceleration A body of mass m starts from rest and reaches velocity v in time t. What is its translational kinetic energy at speed v?

Introduction / Context:
Kinetic energy quantifies the energy of motion. It appears in work–energy calculations, impact analysis, and machine design. This item checks recall and interpretation of the kinetic energy expression for translational motion.

Given Data / Assumptions:

• Mass = m (constant).
• Initial speed = 0 (starts from rest); final speed = v reached in time t.
• We are asked for kinetic energy at speed v, not for the acceleration or distance.

Concept / Approach:
The standard formula for translational kinetic energy is KE = 1/2 * m * v^2. It comes from integrating work done by a net force to increase speed from 0 to v. The time t is incidental here; KE depends only on instantaneous mass and speed, not explicitly on time.

Step-by-Step Solution:
Use work–energy relation: Work done on the particle equals change in kinetic energy.Starting from rest, KE_initial = 0.At speed v, KE_final = 1/2 * m * v^2.Therefore KE = 0.5 * m * v^2.

Verification / Alternative check:
From dynamics: Let constant acceleration be a. With v^2 = 2 * a * s, the work W = Force * distance = m * a * s = m * (v^2 / (2 * s)) * s = 1/2 * m * v^2, matching the formula.

Why Other Options Are Wrong:

• mv^2: misses the factor 1/2.
• mgv^2 and 0.5 mgv^2: insert g (acceleration due to gravity) incorrectly; g is irrelevant to KE unless converting between gravitational potential and kinetic energy.

Common Pitfalls:

• Including g by habit from weight-related problems.
• Confusing power or momentum with kinetic energy.

Final Answer:
0.5 mv2