Answer :
To solve this problem, we will use the formula for kinetic energy \( KE = \frac{1}{2} m v^2 \).
Given:
- Initial velocity (\( v_0 \)) = 2 meters/second
- Initial kinetic energy (\( KE_0 \)) = 40 joules
### Step 1: Determine the mass of the boy
From the initial kinetic energy formula:
[tex]\[ KE_0 = \frac{1}{2} m v_0^2 \][/tex]
Plug in the known values:
[tex]\[ 40 = \frac{1}{2} m \cdot 2^2 \][/tex]
[tex]\[ 40 = \frac{1}{2} m \cdot 4 \][/tex]
[tex]\[ 40 = 2m \][/tex]
Solve for \( m \):
[tex]\[ m = \frac{40}{2} \][/tex]
[tex]\[ m = 20 \, \text{kilograms} \][/tex]
So, the boy weighs 20 kilograms.
### Step 2: Calculate the final velocity
The boy doubles his speed:
[tex]\[ v_f = 2 \times v_0 \][/tex]
[tex]\[ v_f = 2 \times 2 \][/tex]
[tex]\[ v_f = 4 \, \text{meters/second} \][/tex]
### Step 3: Determine the kinetic energy at the faster speed
Using the kinetic energy formula for the final state:
[tex]\[ KE_f = \frac{1}{2} m v_f^2 \][/tex]
Substitute the known values:
[tex]\[ KE_f = \frac{1}{2} \times 20 \times 4^2 \][/tex]
[tex]\[ KE_f = \frac{1}{2} \times 20 \times 16 \][/tex]
[tex]\[ KE_f = 10 \times 16 \][/tex]
[tex]\[ KE_f = 160 \, \text{joules} \][/tex]
So, his kinetic energy at the faster speed is 160 joules.
Therefore, the correct answers are:
- A boy weighing 20 kilograms is riding a skateboard.
- His kinetic energy at the faster speed is 160 joules.
Given:
- Initial velocity (\( v_0 \)) = 2 meters/second
- Initial kinetic energy (\( KE_0 \)) = 40 joules
### Step 1: Determine the mass of the boy
From the initial kinetic energy formula:
[tex]\[ KE_0 = \frac{1}{2} m v_0^2 \][/tex]
Plug in the known values:
[tex]\[ 40 = \frac{1}{2} m \cdot 2^2 \][/tex]
[tex]\[ 40 = \frac{1}{2} m \cdot 4 \][/tex]
[tex]\[ 40 = 2m \][/tex]
Solve for \( m \):
[tex]\[ m = \frac{40}{2} \][/tex]
[tex]\[ m = 20 \, \text{kilograms} \][/tex]
So, the boy weighs 20 kilograms.
### Step 2: Calculate the final velocity
The boy doubles his speed:
[tex]\[ v_f = 2 \times v_0 \][/tex]
[tex]\[ v_f = 2 \times 2 \][/tex]
[tex]\[ v_f = 4 \, \text{meters/second} \][/tex]
### Step 3: Determine the kinetic energy at the faster speed
Using the kinetic energy formula for the final state:
[tex]\[ KE_f = \frac{1}{2} m v_f^2 \][/tex]
Substitute the known values:
[tex]\[ KE_f = \frac{1}{2} \times 20 \times 4^2 \][/tex]
[tex]\[ KE_f = \frac{1}{2} \times 20 \times 16 \][/tex]
[tex]\[ KE_f = 10 \times 16 \][/tex]
[tex]\[ KE_f = 160 \, \text{joules} \][/tex]
So, his kinetic energy at the faster speed is 160 joules.
Therefore, the correct answers are:
- A boy weighing 20 kilograms is riding a skateboard.
- His kinetic energy at the faster speed is 160 joules.
