Final answer:
Using conservation of energy, the mass having 400 J of kinetic energy before impact was dropped from a height of 20.4 m and struck the ground at a speed of 20 m/s.
Step-by-step explanation:
The student's question asks about the final velocity and height from which a 2.0kg mass was dropped if it has 400 J of kinetic energy (KE) before striking the ground, ignoring friction. To solve the mathematical problem completely, we use the principles of conservation of energy. Assuming no energy is lost to air resistance, the potential energy (PE) the mass had at height h is converted to KE just before impact.
First, we can calculate the speed at which it strikes the ground using the KE formula KE = 1/2 * m * v2 where m is mass and v is velocity. Rearranging for velocity, we get:
Substituting the given values,
- v = √(2 * 400 J / 2.0 kg) = √(400 J / kg) = 20 m/s
Next, we determine the height from which it was dropped using the potential energy formula PE = m * g * h, where g is the acceleration due to gravity (approximately 9.81 m/s2). Setting PE equal to the 400 J of KE we have:
- h = KE / (m * g)
- h = 400 J / (2.0 kg * 9.81 m/s2) = 20.4 m
So the correct option is 20.4 m for the height and 20 m/s for the speed at impact.
Therefore, to give you the correct option answer in the final answer, the mass was dropped from a 20.4 m height, and it strikes the ground at a speed of 20 m/s.