Certainly! To calculate the lower and upper bounds of the kinetic energy given the mass and speed with their respective accuracy constraints, follow these steps:
### Step 1: Determine the bounds for mass and speed
First, let's identify the uncertainty in the given values:
1. Mass (m): Given as 35 kg to the nearest 5 kg, the range for the mass can be found by considering ±2.5 kg.
- Lower bound of mass: \( m_{\text{lower}} = 35 - 2.5 = 32.5 \) kg
- Upper bound of mass: \( m_{\text{upper}} = 35 + 2.5 = 37.5 \) kg
2. Speed (v): Given as 5.7 m/s to 2 significant figures, the range for the speed can be found by considering the possible values within the significant figures.
- Lower bound of speed: \( v_{\text{lower}} = 5.65 \) m/s
- Upper bound of speed: \( v_{\text{upper}} = 5.75 \) m/s
### Step 2: Calculate the kinetic energy for the lower and upper bounds
Now, use the kinetic energy formula \( K = \frac{1}{2}mv^2 \) to find the lower and upper bounds for the kinetic energy.
1. Lower bound of kinetic energy (using \( m_{\text{lower}} \) and \( v_{\text{lower}} \)):
[tex]\[
K_{\text{lower}} = \frac{1}{2} \times 32.5 \times (5.65)^2
\][/tex]
Simplifying, we get:
[tex]\[
K_{\text{lower}} \approx 518.7 \, \text{J}
\][/tex]
2. Upper bound of kinetic energy (using \( m_{\text{upper}} \) and \( v_{\text{upper}} \)):
[tex]\[
K_{\text{upper}} = \frac{1}{2} \times 37.5 \times (5.75)^2
\][/tex]
Simplifying, we get:
[tex]\[
K_{\text{upper}} \approx 619.9 \, \text{J}
\][/tex]
### Summary of results:
- Mass Lower Bound: 32.5 kg
- Mass Upper Bound: 37.5 kg
- Speed Lower Bound: 5.65 m/s
- Speed Upper Bound: 5.75 m/s
- Kinetic Energy Lower Bound: 518.7 J
- Kinetic Energy Upper Bound: 619.9 J
Hence, the lower and upper bounds of the kinetic energy of the object are approximately 518.7 J and 619.9 J, respectively, when rounded to 1 decimal place.
