A 4 kg object is moving horizontally with a speed of 5 m/s. To increase its speed to 10 m/s, how much net work must be done on the object?

Introduction / Context:

This question is a straightforward application of the work energy theorem in mechanics. The theorem states that the net work done on an object equals the change in its kinetic energy. By comparing the initial and final kinetic energies of a moving object, we can determine how much work is required to increase its speed from one value to another. This concept is very useful in physics, engineering, and real life calculations, such as estimating the energy needed for acceleration in vehicles.

Given Data / Assumptions:

• Mass of the object, m = 4 kg.
• Initial speed, v1 = 5 m/s.
• Final speed, v2 = 10 m/s.
• The motion is horizontal and friction or other energy losses are neglected; only net work is considered.

Concept / Approach:

The kinetic energy KE of a body of mass m moving with speed v is KE = (1/2) * m * v^2. The work energy theorem states that net work done, W_net, equals the change in kinetic energy: W_net = KE_final - KE_initial. To solve the problem, we first compute initial and final kinetic energies, then subtract to find the work done. Careful substitution and correct squaring of speeds are essential to arrive at the correct numerical answer.

Step-by-Step Solution:

Verification / Alternative check:

We can also reason qualitatively: kinetic energy depends on the square of the speed. Doubling the speed from 5 m/s to 10 m/s increases v^2 by a factor of 4 (from 25 to 100). Therefore, kinetic energy should also increase by a factor of 4, assuming mass is constant. Initial KE is 50 J, and four times that is 200 J, which matches the calculated final KE. The difference, 200 - 50 = 150 J, is the needed work, confirming the result.

Why Other Options Are Wrong:

• 100 J: This might arise from incorrectly using linear proportionality with speed instead of speed squared, or from calculation mistakes.
• 50 J: This equals the initial kinetic energy and would only raise the energy from 0 to 50 J, not from 50 J to 200 J.
• 75 J: This is half the correct answer and reflects a miscalculation or partial change in kinetic energy.

Common Pitfalls:

A typical error is to treat kinetic energy as proportional to speed rather than speed squared, leading to incorrect scaling when speed is doubled. Another common mistake is mishandling the square during calculation, such as squaring the mass instead of the speed. Finally, students sometimes forget to subtract initial kinetic energy and incorrectly equate work done with final kinetic energy only. Carefully applying the formula KE = (1/2) * m * v^2 and the work energy theorem helps avoid these issues.

Final Answer:

The required net work is 150 J.