To solve the problem step by step, we need to calculate the kinetic energy and potential energy of the stone after 1 second of being thrown vertically upward.
### Step 1: Calculate the height after 1 second
We can use the equation of motion to find the height (S) after 1 second:
\[ S = ut + \frac{1}{2} a t^2 \]
Where:
- \( u = 20 \, \text{m/s} \) (initial velocity)
- \( t = 1 \, \text{s} \) (time)
- \( a = -g = -10 \, \text{m/s}^2 \) (acceleration due to gravity, negative because it acts downward)
Substituting the values:
\[ S = (20 \, \text{m/s})(1 \, \text{s}) + \frac{1}{2} (-10 \, \text{m/s}^2)(1 \, \text{s})^2 \]
\[ S = 20 \, \text{m} - 5 \, \text{m} \]
\[ S = 15 \, \text{m} \]
### Step 2: Calculate the kinetic energy after 1 second
To find the kinetic energy (KE) after 1 second, we first need to calculate the final velocity (v) at that time using:
\[ v = u + at \]
Substituting the values:
\[ v = 20 \, \text{m/s} + (-10 \, \text{m/s}^2)(1 \, \text{s}) \]
\[ v = 20 \, \text{m/s} - 10 \, \text{m/s} \]
\[ v = 10 \, \text{m/s} \]
Now, we can calculate the kinetic energy using the formula:
\[ KE = \frac{1}{2} mv^2 \]
Where:
- \( m = 3 \, \text{kg} \)
- \( v = 10 \, \text{m/s} \)
Substituting the values:
\[ KE = \frac{1}{2} (3 \, \text{kg}) (10 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} (3) (100) \]
\[ KE = \frac{300}{2} \]
\[ KE = 150 \, \text{J} \]
### Step 3: Calculate the potential energy after 1 second
The potential energy (PE) can be calculated using the formula:
\[ PE = mgh \]
Where:
- \( m = 3 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( h = 15 \, \text{m} \) (height calculated in Step 1)
Substituting the values:
\[ PE = (3 \, \text{kg})(10 \, \text{m/s}^2)(15 \, \text{m}) \]
\[ PE = 3 \times 10 \times 15 \]
\[ PE = 450 \, \text{J} \]
### Summary of Results
- Kinetic Energy after 1 second: \( KE = 150 \, \text{J} \)
- Potential Energy after 1 second: \( PE = 450 \, \text{J} \)
