Question
A 2-kg object is dropped from a height of 10 meters. Using the formula _____ (use g=10m/s2), the potential energy is calculated as _____. If the object strikes the ground with a speed of 5m/s, then using the formula _____, the kinetic energy is 100 joules. This requires that _____ of energy is turned into ______ and everntually released to its surroundings.__
A. k=1/2 mv*2
B.U=mgh
C.100 joules
D.heat 200 joules
A. k=1/2 mv*2
B.U=mgh
C.100 joules
D.heat 200 joules
Answers
GPT-4o mini
To fill in the blanks in your statements, we can break down the problem using the formulas for potential energy (PE) and kinetic energy (KE), as well as facts about conservation of energy:
1. The formula for calculating potential energy is **B. U = mgh**.
2. For an object of mass 2 kg dropped from a height of 10 meters with \( g = 10 \, \text{m/s}^2 \):
\[
U = mgh = 2 \, \text{kg} \times 10 \, \text{m/s}^2 \times 10 \, \text{m} = 200 \, \text{joules}.
\]
Therefore, the potential energy calculated is **200 joules**.
3. If the object strikes the ground with a speed of 5 m/s, the formula for kinetic energy is **A. k = 1/2 mv²**. Calculating kinetic energy:
\[
k = \frac{1}{2}mv^2 = \frac{1}{2} \times 2 \, \text{kg} \times (5 \, \text{m/s})^2 = \frac{1}{2} \times 2 \times 25 = 25 \, \text{joules}.
\]
Since you mentioned that kinetic energy is 100 joules, this seems to contradict my calculation based on the given speed. The speed must be higher if the KE is truly 100 joules.
4. Potential Energy before dropping is 200 joules. When it converts to other forms:
Assuming all potential energy was converted to kinetic energy and other forms (like heat):
This requires that **D. 200 joules** of energy is turned into **D. heat** and eventually released to its surroundings.
So the filled statement reads as:
A 2-kg object is dropped from a height of 10 meters. Using the formula **B. U = mgh**, the potential energy is calculated as **200 joules**. If the object strikes the ground with a speed of 5m/s, then using the formula **A. k = 1/2 mv²**, the kinetic energy is **25 joules**. This requires that **200 joules** of energy is turned into **heat** and eventually released to its surroundings.
1. The formula for calculating potential energy is **B. U = mgh**.
2. For an object of mass 2 kg dropped from a height of 10 meters with \( g = 10 \, \text{m/s}^2 \):
\[
U = mgh = 2 \, \text{kg} \times 10 \, \text{m/s}^2 \times 10 \, \text{m} = 200 \, \text{joules}.
\]
Therefore, the potential energy calculated is **200 joules**.
3. If the object strikes the ground with a speed of 5 m/s, the formula for kinetic energy is **A. k = 1/2 mv²**. Calculating kinetic energy:
\[
k = \frac{1}{2}mv^2 = \frac{1}{2} \times 2 \, \text{kg} \times (5 \, \text{m/s})^2 = \frac{1}{2} \times 2 \times 25 = 25 \, \text{joules}.
\]
Since you mentioned that kinetic energy is 100 joules, this seems to contradict my calculation based on the given speed. The speed must be higher if the KE is truly 100 joules.
4. Potential Energy before dropping is 200 joules. When it converts to other forms:
Assuming all potential energy was converted to kinetic energy and other forms (like heat):
This requires that **D. 200 joules** of energy is turned into **D. heat** and eventually released to its surroundings.
So the filled statement reads as:
A 2-kg object is dropped from a height of 10 meters. Using the formula **B. U = mgh**, the potential energy is calculated as **200 joules**. If the object strikes the ground with a speed of 5m/s, then using the formula **A. k = 1/2 mv²**, the kinetic energy is **25 joules**. This requires that **200 joules** of energy is turned into **heat** and eventually released to its surroundings.