Get detailed and accurate answers to your questions on IDNLearn.com. Our Q&A platform offers detailed and trustworthy answers to ensure you have the information you need.
Sagot :
Let's break down the problem step by step and solve it systematically.
### Step 1: Mass of the Cart
We are given the following information:
- A constant net force [tex]\( F_{\text{net}} = 200 \)[/tex] N is exerted.
- The cart accelerates from rest, so the initial velocity [tex]\( u = 0 \)[/tex] m/s.
- The final velocity [tex]\( v = 4 \)[/tex] m/s.
- We need to find the mass of the cart [tex]\( m_{\text{cart}} \)[/tex].
#### Step 1.1: Calculate the acceleration of the cart
Use the formula for acceleration:
[tex]\[ a = \frac{v - u}{t} \][/tex]
Assuming the time [tex]\( t \)[/tex] taken for this change in velocity is 1 second (for simplicity), the acceleration [tex]\( a \)[/tex] can be calculated as:
[tex]\[ a = \frac{4 \text{ m/s} - 0 \text{ m/s}}{1 \text{ s}} = 4 \text{ m/s}^2 \][/tex]
#### Step 1.2: Calculate the mass of the cart
Use Newton's second law, which states that force is the product of mass and acceleration:
[tex]\[ F = ma \][/tex]
Rearranging for mass:
[tex]\[ m = \frac{F}{a} \][/tex]
Substitute the given values:
[tex]\[ m_{\text{cart}} = \frac{200 \text{ N}}{4 \text{ m/s}^2} = 50 \text{ kg} \][/tex]
So, the mass of the cart is 50 kg.
### Step 2: Acceleration of the Ball
We are given the following information:
- The mass of the ball [tex]\( m_{\text{ball}} = 0.40 \)[/tex] kg.
- A force [tex]\( F_{\text{ball}} = 200 \)[/tex] N is exerted on the ball.
#### Step 2.1: Calculate the acceleration of the ball
Again, use Newton's second law:
[tex]\[ F_{\text{ball}} = m_{\text{ball}} \cdot a_{\text{ball}} \][/tex]
Rearrange to find the acceleration:
[tex]\[ a_{\text{ball}} = \frac{F_{\text{ball}}}{m_{\text{ball}}} \][/tex]
Substitute the given values:
[tex]\[ a_{\text{ball}} = \frac{200 \text{ N}}{0.40 \text{ kg}} = 500 \text{ m/s}^2 \][/tex]
So, the acceleration of the ball is 500 m/s².
### Conclusion
1. The mass of the cart is 50 kg.
2. The acceleration of the ball is 500 m/s².
### Step 1: Mass of the Cart
We are given the following information:
- A constant net force [tex]\( F_{\text{net}} = 200 \)[/tex] N is exerted.
- The cart accelerates from rest, so the initial velocity [tex]\( u = 0 \)[/tex] m/s.
- The final velocity [tex]\( v = 4 \)[/tex] m/s.
- We need to find the mass of the cart [tex]\( m_{\text{cart}} \)[/tex].
#### Step 1.1: Calculate the acceleration of the cart
Use the formula for acceleration:
[tex]\[ a = \frac{v - u}{t} \][/tex]
Assuming the time [tex]\( t \)[/tex] taken for this change in velocity is 1 second (for simplicity), the acceleration [tex]\( a \)[/tex] can be calculated as:
[tex]\[ a = \frac{4 \text{ m/s} - 0 \text{ m/s}}{1 \text{ s}} = 4 \text{ m/s}^2 \][/tex]
#### Step 1.2: Calculate the mass of the cart
Use Newton's second law, which states that force is the product of mass and acceleration:
[tex]\[ F = ma \][/tex]
Rearranging for mass:
[tex]\[ m = \frac{F}{a} \][/tex]
Substitute the given values:
[tex]\[ m_{\text{cart}} = \frac{200 \text{ N}}{4 \text{ m/s}^2} = 50 \text{ kg} \][/tex]
So, the mass of the cart is 50 kg.
### Step 2: Acceleration of the Ball
We are given the following information:
- The mass of the ball [tex]\( m_{\text{ball}} = 0.40 \)[/tex] kg.
- A force [tex]\( F_{\text{ball}} = 200 \)[/tex] N is exerted on the ball.
#### Step 2.1: Calculate the acceleration of the ball
Again, use Newton's second law:
[tex]\[ F_{\text{ball}} = m_{\text{ball}} \cdot a_{\text{ball}} \][/tex]
Rearrange to find the acceleration:
[tex]\[ a_{\text{ball}} = \frac{F_{\text{ball}}}{m_{\text{ball}}} \][/tex]
Substitute the given values:
[tex]\[ a_{\text{ball}} = \frac{200 \text{ N}}{0.40 \text{ kg}} = 500 \text{ m/s}^2 \][/tex]
So, the acceleration of the ball is 500 m/s².
### Conclusion
1. The mass of the cart is 50 kg.
2. The acceleration of the ball is 500 m/s².
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.