A body of mass 4.0 kilogram is initially at rest. Under the action of a constant unbalanced force it gains a speed of 5 metre per second. How much work is done by the force on the body?

Difficulty: Medium

Correct Answer: 50 joule

Explanation:


Introduction / Context:

Work and kinetic energy are closely related in mechanics. When a net force acts on a body and changes its speed, the work done by that force is equal to the change in the kinetic energy of the body. This question uses that relationship to find the work done when a body is accelerated from rest to a certain speed.


Given Data / Assumptions:

  • Mass m of the body is 4.0 kilogram.
  • Initial velocity u is 0 metre per second because the body is at rest.
  • Final velocity v is 5 metre per second.
  • The applied force is constant and acts along the direction of motion.


Concept / Approach:

The work energy theorem states that the net work done on an object equals the change in its kinetic energy. Mathematically, W = KE final minus KE initial. Kinetic energy is given by KE = (1 / 2) * m * v^2. Using the initial and final velocities, we can calculate the change in kinetic energy and thus the work done by the force.


Step-by-Step Solution:

Step 1: Compute initial kinetic energy: KE initial = (1 / 2) * m * u^2. Since u = 0, KE initial = 0. Step 2: Compute final kinetic energy: KE final = (1 / 2) * m * v^2 = (1 / 2) * 4 * 5^2. Step 3: Calculate v^2 = 5^2 = 25, so KE final = (1 / 2) * 4 * 25 = 2 * 25 = 50 joule. Step 4: Use work energy theorem: W = KE final minus KE initial = 50 minus 0 = 50 joule. So the work done is 50 joule.


Verification / Alternative check:

The answer can be checked by estimating. A 4 kilogram mass moving at 5 metre per second is not extremely fast, so an energy of tens of joule is reasonable. The calculation 0.5 * 4 * 25 naturally gives 50, so the arithmetic is straightforward and consistent.


Why Other Options Are Wrong:

Option B: 30 joule does not match the computed change in kinetic energy and likely arises from a random mistake.

Option C: 20 joule could result from using an incorrect formula such as m * v instead of (1 / 2) * m * v^2.

Option D: 40 joule might come from miscalculating 0.5 * 4 * 5^2 as 40 instead of 50, showing a multiplication error.


Common Pitfalls:

Students sometimes forget that work depends on change in kinetic energy, not just on final energy. Others may incorrectly square the velocity or misplace decimal points. Keeping the formula W = (1 / 2) * m * v^2 minus (1 / 2) * m * u^2 in mind and performing each step carefully helps avoid such mistakes.


Final Answer:

The work done by the force on the body is 50 joule.

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