Question
Wyatt pulls on a sled that his friend is sitting on until the sled just begins to move. His friend and the sled have a combined mass of 80 kg. The image below shows a free-body diagram of all the forces acting on Wyatt’s friend and the sled.
What is the acceleration of Wyatt’s friend and the sled?
A.
10 m/s2
B.
3 m/s2
C.
8 m/s2
D.
5 m/s2
What is the acceleration of Wyatt’s friend and the sled?
A.
10 m/s2
B.
3 m/s2
C.
8 m/s2
D.
5 m/s2
Answers
GPT-4o mini
To determine the acceleration of Wyatt's friend and the sled, we can use Newton’s second law of motion, which states that \( F = ma \), where \( F \) is the net force acting on an object, \( m \) is the mass, and \( a \) is the acceleration.
First, we need to identify the forces acting on the system. When Wyatt pulls the sled, there will be a force exerted by Wyatt (let's call it \( F_{\text{pull}} \)) and the force of friction opposing the motion (let's call it \( F_{\text{friction}} \)).
Since we don’t have the exact values for these forces, but we know the sled just begins to move, the pulling force equals the force of static friction right at the moment of motion initiation.
Given that the combined mass of the friend and the sled (\( m \)) is 80 kg, and if we assume that Wyatt is able to pull with a force that just overcomes static friction, we can find the acceleration once the sled is already moving.
**However**, without an explicit value of the pulling force or the friction force, we can’t directly calculate \( a \).
From the choices given:
- A. \( 10 \, m/s^2 \)
- B. \( 3 \, m/s^2 \)
- C. \( 8 \, m/s^2 \)
- D. \( 5 \, m/s^2 \)
If we assume the gravitational force acting on the mass is \( F_{\text{gravity}} = m \cdot g = 80 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 784 \, \text{N} \), and if static friction is close but not overwhelming (most cases sees a static friction coefficient range typical around 0.2 to 0.5 for sleds), the force needed to reach motion might be substantially lesser than the total weight pulling down.
If this was the case, we might see an acceleration that’s calculated directly from an order of magnitude.
Let’s consider if any typical coefficient of friction (like 0.2), gives \( F_{\text{friction}} \) = \( \mu * N\) = \( 0.3 * 784 = 235.2 \, N\) leading \( F_{\text{pull}} - F_{\text{friction}} = ma\).
However, the problem statement does not appear to give data on required forces directly.
Since often in physics problems with options provided, a more common setup implies the simpler results occur at lower orders of magnitude unless extreme values are given, likely leaning towards either D or B could be a better assumption.
From the situation and without other numerical force data, **Option D (5 m/s²)** would correspond most commonly with a pull capability exceeding usual friction ranges for setups like sled movements. Thus the acceleration could realistically be estimated as nearby.
**Therefore, answer likely would be: D. 5 m/s².**
First, we need to identify the forces acting on the system. When Wyatt pulls the sled, there will be a force exerted by Wyatt (let's call it \( F_{\text{pull}} \)) and the force of friction opposing the motion (let's call it \( F_{\text{friction}} \)).
Since we don’t have the exact values for these forces, but we know the sled just begins to move, the pulling force equals the force of static friction right at the moment of motion initiation.
Given that the combined mass of the friend and the sled (\( m \)) is 80 kg, and if we assume that Wyatt is able to pull with a force that just overcomes static friction, we can find the acceleration once the sled is already moving.
**However**, without an explicit value of the pulling force or the friction force, we can’t directly calculate \( a \).
From the choices given:
- A. \( 10 \, m/s^2 \)
- B. \( 3 \, m/s^2 \)
- C. \( 8 \, m/s^2 \)
- D. \( 5 \, m/s^2 \)
If we assume the gravitational force acting on the mass is \( F_{\text{gravity}} = m \cdot g = 80 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 784 \, \text{N} \), and if static friction is close but not overwhelming (most cases sees a static friction coefficient range typical around 0.2 to 0.5 for sleds), the force needed to reach motion might be substantially lesser than the total weight pulling down.
If this was the case, we might see an acceleration that’s calculated directly from an order of magnitude.
Let’s consider if any typical coefficient of friction (like 0.2), gives \( F_{\text{friction}} \) = \( \mu * N\) = \( 0.3 * 784 = 235.2 \, N\) leading \( F_{\text{pull}} - F_{\text{friction}} = ma\).
However, the problem statement does not appear to give data on required forces directly.
Since often in physics problems with options provided, a more common setup implies the simpler results occur at lower orders of magnitude unless extreme values are given, likely leaning towards either D or B could be a better assumption.
From the situation and without other numerical force data, **Option D (5 m/s²)** would correspond most commonly with a pull capability exceeding usual friction ranges for setups like sled movements. Thus the acceleration could realistically be estimated as nearby.
**Therefore, answer likely would be: D. 5 m/s².**