A body of mass 2 kg is thrown vertically upward with an initial velocity of 20 m/s. Taking g = 10 m/s^2, what will be its kinetic energy after 2 seconds of upward motion?

Introduction / Context:
Projectile motion and vertical throws are common topics in basic mechanics. When a body is thrown straight up, gravity slows it down until its velocity becomes zero at the highest point, and then it falls back. This question connects kinematics with kinetic energy by asking what the kinetic energy of a body is after a certain time during its ascent. Understanding how velocity changes under constant acceleration due to gravity and how kinetic energy depends on velocity is very important for many physics problems in examinations.

Given Data / Assumptions:
• Mass m = 2 kg. • Initial velocity u = 20 m/s upward. • Acceleration due to gravity g = 10 m/s^2 acting downward. • Time of motion considered t = 2 s after the body is thrown. • Air resistance is neglected and motion is vertical.

Concept / Approach:
The motion is uniformly accelerated under gravity. For an object thrown upward, its velocity decreases linearly with time according to the equation v = u - g * t, where v is the velocity at time t. Once we find the velocity after 2 seconds, we can calculate the kinetic energy using the standard expression for translational kinetic energy, KE = (1 / 2) * m * v^2. If the velocity becomes zero at that instant, the kinetic energy will also become zero, because kinetic energy depends on the square of speed.

Step-by-Step Solution:
Step 1: Use the kinematic relation v = u - g * t for upward motion. Step 2: Substitute u = 20 m/s, g = 10 m/s^2, and t = 2 s. Step 3: Compute v = 20 - 10 * 2 = 20 - 20 = 0 m/s. Step 4: Use the kinetic energy formula KE = (1 / 2) * m * v^2. Step 5: Substitute m = 2 kg and v = 0 m/s to get KE = (1 / 2) * 2 * 0^2 = 0 J. Step 6: Therefore, the kinetic energy after 2 seconds is 0 J.

Verification / Alternative check:
We can also reason physically. The time to reach the highest point is t_top = u / g = 20 / 10 = 2 s. This means that exactly after 2 seconds, the body reaches its maximum height, and at that instant its velocity becomes zero before it starts descending. At the highest point in a vertical throw, all the initial kinetic energy has been converted into gravitational potential energy. Since kinetic energy is zero when velocity is zero, the answer of 0 J is fully consistent with the energy viewpoint as well.

Why Other Options Are Wrong:
Option b (100 J): This would correspond to a non zero velocity and does not match the correct v = 0 m/s at t = 2 s. Option c (200 J): This value is equal to the initial kinetic energy (1 / 2 * 2 * 20^2 = 400 J) divided by 2, but there is no reason for the energy to be half at this particular time. Option d (400 J): This is actually the initial kinetic energy at t = 0 s, not the kinetic energy after 2 seconds. Option e (20 J): This is an arbitrary small value that does not follow from the kinematic or energy calculations.

Common Pitfalls:
A common error is to forget the sign in the kinematic equation and write v = u + g * t for upward motion, which would give an incorrect non zero velocity. Some learners also directly use displacement formulas instead of focusing on velocity, which can lead to confusion. It is also easy to mistakenly assume that kinetic energy decreases linearly with time, but in reality it depends on the square of velocity. Always calculate the velocity at the required time using correct sign conventions and then substitute into the energy formula.

Final Answer:
The kinetic energy of the body after 2 seconds is 0 J.