Calculating Kinetic Energy of a Body Thrown Upward

This problem asks us to determine the kinetic energy of a body after it has been thrown vertically upward for a specific duration. We are given the body's mass, initial velocity, the time elapsed, and the value of acceleration due to gravity.

Understanding the Problem: Motion Under Gravity

When a body is thrown upward, its motion is influenced by gravity, which acts downwards. This causes the body to decelerate as it moves upwards. To find the kinetic energy after a certain time, we first need to determine the body's velocity at that specific moment using kinematic equations for motion under constant acceleration (gravity).

Given Information

Let's list the information provided:

Step-by-Step Calculation

Step 1: Calculate the velocity after 2 seconds

We use the first equation of motion, which relates initial velocity, final velocity, acceleration, and time. Since the body is moving upward and gravity acts downward, the acceleration (\(a\)) is negative \(g\).

The equation is:

\(v = u + at\)

Here, \(u = 20 \, \text{m/s}\), \(a = -g = -10 \, \text{m/s}^2\), and \(t = 2 \, \text{s}\).

Substitute these values into the equation:

\(v = 20 \, \text{m/s} + (-10 \, \text{m/s}^2)(2 \, \text{s})\)

\(v = 20 \, \text{m/s} - 20 \, \text{m/s}\)

\(v = 0 \, \text{m/s}\)

So, after 2 seconds, the velocity of the body is 0 m/s.

Step 2: Calculate the Kinetic Energy

Kinetic energy (KE) is the energy possessed by a body due to its motion. It is given by the formula:

\(KE = \frac{1}{2}mv^2\)

Here, \(m = 2 \, \text{kg}\) and the velocity \(v\) after 2 seconds is \(0 \, \text{m/s}\).

Substitute these values into the kinetic energy formula:

\(KE = \frac{1}{2}(2 \, \text{kg})(0 \, \text{m/s})^2\)

\(KE = \frac{1}{2}(2)(0)\) Joules

\(KE = 1 \times 0\) Joules

\(KE = 0\) Joules

Therefore, after 2 seconds, the kinetic energy of the body is 0 J.

The result indicates that at the 2-second mark, the body momentarily stops before starting to fall back down. This suggests that 2 seconds is the time it takes for the body to reach its maximum height in this scenario.

Kinetic Energy Calculation Summary

• Initial velocity \(u = 20\) m/s
• Acceleration \(a = -10\) m/s\(^2\)
• Time \(t = 2\) s
• Final velocity \(v = u + at = 20 + (-10)(2) = 0\) m/s
• Mass \(m = 2\) kg
• Kinetic Energy \(KE = \frac{1}{2}mv^2 = \frac{1}{2}(2)(0)^2 = 0\) J

Conclusion

After 2 seconds, the body's velocity is 0 m/s. Consequently, its kinetic energy is also 0 J.

Revision Table - Key Concepts

Additional Information - Projectile Motion

This problem is an example of one-dimensional projectile motion (vertical motion). When an object is thrown upward:

• Its velocity decreases as it rises due to the downward acceleration of gravity.
• At its maximum height, its instantaneous velocity is zero. This is the point where it stops moving up and begins to fall down.
• The time taken to reach the maximum height is given by \(t_{max} = \frac{u}{g}\) (where \(u\) is initial upward velocity and \(g\) is the magnitude of gravity). In this problem, \(t_{max} = \frac{20}{10} = 2\) seconds, confirming our finding that the velocity is zero at 2 seconds.
• The motion going up is symmetrical to the motion coming down in the absence of air resistance (which is typically assumed in such problems).

The total mechanical energy (sum of kinetic and potential energy) is conserved throughout the motion, assuming no air resistance. As the body rises, kinetic energy is converted into potential energy, and as it falls, potential energy is converted back into kinetic energy.