A body of mass 4 kg has a kinetic energy of 72 J. Its speed is:

Correct option is D

The correct answer is (D) 6 m/s

Explanation:
• Kinetic energy represents the scalar energy possessed by an object due to its motion. The relationship between the mass (m) of an object, its linear speed (v), and its kinetic energy (KE) is defined by the formula: $$KE = \frac{1}{2}mv^2$$​.
• We can rearrange this equation to solve explicitly for the speed (v). Multiplying both sides by 2 gives $$2 \times KE = mv^2$$​, and dividing by mass yields $$v^2 = \frac{2 \times KE}{m}$$​. Taking the square root of both sides gives the expression: $$v = \sqrt{\frac{2 \times KE}{m}}$$​.
• Now, we substitute the values provided in the problem:
- Mass (m) =$$4\text{ kg}$$​​
- Kinetic Energy (KE) = $$72\text{ Joules}$$​​
• Substituting these values into the rearranged equation: $$v^2 = \frac{2 \times 72}{4}$$​. Simplifying the fraction: $$v^2 = \frac{144}{4} = 36$$​.
• To find the final speed, we calculate the square root of 36: $$v = \sqrt{36} = 6\text{ m/s}$$​. Thus, the speed of the body is exactly $$6\text{ meters per second}$$​.

Information Booster:
• The Work-Energy Theorem states that the net work done on an object is equal to the change in its kinetic energy ($$W = \Delta KE$$​). In this case, to accelerate a $$4\text{ kg}$$​ object from rest to a speed of $$6\text{ m/s}$$​, a net work of exactly $$72\text{ J}$$​ must be performed on it.
• Kinetic energy can also be expressed in terms of linear momentum (p = mv) using the equation $$KE = \frac{p^2}{2m}$$​. Using our calculated speed, the momentum of this body is $$p = 4\text{ kg} \times 6\text{ m/s} = 24\text{ kg}\cdot\text{m/s}$$​. Substituting this back gives $$KE = \frac{24^2}{2 \times 4} = \frac{576}{8} = 72\text{ J}$$​, which confirms our calculation.

Additional Knowledge:
• 4 m/s (Option A): If the body were traveling at this speed, its kinetic energy would be significantly lower: $$KE = \frac{1}{2} \times 4 \times 4^2 = 2 \times 16 = 32\text{ J}$$​.
• 3 m/s (Option B): At a speed of $$3\text{ m/s}$$​, the kinetic energy would drop to: $$KE = \frac{1}{2} \times 4 \times 3^2 = 2 \times 9 = 18\text{ J}$$​.
• 9 m/s (Option C): If the body accelerated to $$9\text{ m/s}$$​, its kinetic energy would increase substantially to: $$KE = \frac{1}{2} \times 4 \times 9^2 = 2 \times 81 = 162\text{ J}$$​.