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.__
Answers
GPT-4o mini
To fill in the blanks in your statements:
1. The formula for potential energy (PE) is \( PE = mgh \), where:
- \( m \) is the mass of the object (2 kg),
- \( g \) is the acceleration due to gravity (10 m/s²),
- \( h \) is the height from which the object is dropped (10 m).
Using these values, the potential energy is calculated as:
\[
PE = 2 \, \text{kg} \times 10 \, \text{m/s}^2 \times 10 \, \text{m} = 200 \, \text{joules}.
\]
2. The object strikes the ground with a speed of 5 m/s. The formula for kinetic energy (KE) is \( KE = \frac{1}{2} mv^2 \), where:
- \( m \) is the mass of the object (2 kg),
- \( v \) is the speed of the object just before impact (5 m/s).
Using these values, the kinetic energy is calculated as:
\[
KE = \frac{1}{2} \times 2 \, \text{kg} \times (5 \, \text{m/s})^2 = \frac{1}{2} \times 2 \times 25 = 25 \, \text{joules}.
\]
(Note: You mentioned 100 joules, but with the values given, it calculates to 25 joules using \( 5 \, \text{m/s} \). Please verify the speed.)
3. This requires that **some amount** of energy is turned into **kinetic energy** and eventually released to its surroundings.
Let's summarize everything:
A 2-kg object is dropped from a height of 10 meters. Using the formula **\( PE = mgh \)**, the potential energy is calculated as **200 joules**. If the object strikes the ground with a speed of 5 m/s, then using the formula **\( KE = \frac{1}{2}mv^2 \)**, the kinetic energy is **25 joules**. This requires that **some amount** of energy is turned into **kinetic energy** and eventually released to its surroundings.
1. The formula for potential energy (PE) is \( PE = mgh \), where:
- \( m \) is the mass of the object (2 kg),
- \( g \) is the acceleration due to gravity (10 m/s²),
- \( h \) is the height from which the object is dropped (10 m).
Using these values, the potential energy is calculated as:
\[
PE = 2 \, \text{kg} \times 10 \, \text{m/s}^2 \times 10 \, \text{m} = 200 \, \text{joules}.
\]
2. The object strikes the ground with a speed of 5 m/s. The formula for kinetic energy (KE) is \( KE = \frac{1}{2} mv^2 \), where:
- \( m \) is the mass of the object (2 kg),
- \( v \) is the speed of the object just before impact (5 m/s).
Using these values, the kinetic energy is calculated as:
\[
KE = \frac{1}{2} \times 2 \, \text{kg} \times (5 \, \text{m/s})^2 = \frac{1}{2} \times 2 \times 25 = 25 \, \text{joules}.
\]
(Note: You mentioned 100 joules, but with the values given, it calculates to 25 joules using \( 5 \, \text{m/s} \). Please verify the speed.)
3. This requires that **some amount** of energy is turned into **kinetic energy** and eventually released to its surroundings.
Let's summarize everything:
A 2-kg object is dropped from a height of 10 meters. Using the formula **\( PE = mgh \)**, the potential energy is calculated as **200 joules**. If the object strikes the ground with a speed of 5 m/s, then using the formula **\( KE = \frac{1}{2}mv^2 \)**, the kinetic energy is **25 joules**. This requires that **some amount** of energy is turned into **kinetic energy** and eventually released to its surroundings.