A ball, with a mass of 5 kg, acceelerates in the +a direction with a magnitude of 5 m/s? and in the -y direction with a magnitude of 2 m/s2. What is the acceleration vector that describes this motion? What is the net force on this ball?

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Chapter3: Vectors
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Problem 13PQ: In Chapter 5, you will study a very important vector, force. Each case in Figure P3.13 shows an...
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**Problem Statement:**

A ball, with a mass of 5 kg, accelerates in the \(+x\) direction with a magnitude of 5 m/s\(^2\) and in the \(-y\) direction with a magnitude of 2 m/s\(^2\). 

1. What is the acceleration vector that describes this motion? 
2. What is the net force on this ball?

**Solution Explanation:**

To find the acceleration vector, we combine the accelerations in the \(x\) and \(y\) directions:

- The acceleration in the \(x\)-direction is \(+5 \, \text{m/s}^2\).
- The acceleration in the \(y\)-direction is \(-2 \, \text{m/s}^2\).

Thus, the acceleration vector \(\mathbf{a}\) is \(\mathbf{a} = (5 \, \text{m/s}^2) \, \mathbf{i} + (-2 \, \text{m/s}^2) \, \mathbf{j}\).

To find the net force, we use Newton's second law, \(\mathbf{F} = m \mathbf{a}\), where \(m = 5 \, \text{kg}\).

Substitute the given values:
\[
\mathbf{F} = 5 \, \text{kg} \cdot \left[(5 \, \text{m/s}^2) \, \mathbf{i} + (-2 \, \text{m/s}^2) \, \mathbf{j}\right]
\]

This results in:
\[
\mathbf{F} = (25 \, \text{N}) \, \mathbf{i} + (-10 \, \text{N}) \, \mathbf{j}
\]

Therefore, the net force on the ball is \(\mathbf{F} = 25 \, \text{N} \, \mathbf{i} - 10 \, \text{N} \, \mathbf{j}\).
Transcribed Image Text:**Problem Statement:** A ball, with a mass of 5 kg, accelerates in the \(+x\) direction with a magnitude of 5 m/s\(^2\) and in the \(-y\) direction with a magnitude of 2 m/s\(^2\). 1. What is the acceleration vector that describes this motion? 2. What is the net force on this ball? **Solution Explanation:** To find the acceleration vector, we combine the accelerations in the \(x\) and \(y\) directions: - The acceleration in the \(x\)-direction is \(+5 \, \text{m/s}^2\). - The acceleration in the \(y\)-direction is \(-2 \, \text{m/s}^2\). Thus, the acceleration vector \(\mathbf{a}\) is \(\mathbf{a} = (5 \, \text{m/s}^2) \, \mathbf{i} + (-2 \, \text{m/s}^2) \, \mathbf{j}\). To find the net force, we use Newton's second law, \(\mathbf{F} = m \mathbf{a}\), where \(m = 5 \, \text{kg}\). Substitute the given values: \[ \mathbf{F} = 5 \, \text{kg} \cdot \left[(5 \, \text{m/s}^2) \, \mathbf{i} + (-2 \, \text{m/s}^2) \, \mathbf{j}\right] \] This results in: \[ \mathbf{F} = (25 \, \text{N}) \, \mathbf{i} + (-10 \, \text{N}) \, \mathbf{j} \] Therefore, the net force on the ball is \(\mathbf{F} = 25 \, \text{N} \, \mathbf{i} - 10 \, \text{N} \, \mathbf{j}\).
Expert Solution
Step 1

Given Data :

The mass of the ball is given as Mb = 5 Kg.

The acceleration of the ball in the x-direction is given as ax=5 ms2

The acceleration of the ball in the y-direction is given as ay=-2 ms2

The negative sign indicates that the ball is moving in the negative y-direction.

 

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