Questions: Two horizontal forces F1 and F2 act on a 4.00 kg disk that slides over frictionless ice, on which an xy coordinate system is laid out. Force F1 is in the positive x direction and has a magnitude of 7.70 N. Force F2 has a magnitude of 9.10 N. The figure here gives the x component vx of the velocity of the disk as a function of time t during the sliding. What is the angle between the constant directions of forces F1 and F2?

Two horizontal forces F1 and F2 act on a 4.00 kg disk that slides over frictionless ice, on which an xy coordinate system is laid out. Force F1 is in the positive x direction and has a magnitude of 7.70 N. Force F2 has a magnitude of 9.10 N. The figure here gives the x component vx of the velocity of the disk as a function of time t during the sliding. What is the angle between the constant directions of forces F1 and F2?

Transcript text: Question 1 of 5 -/5 Current Attempt in Progress Two horizontal forces $\vec{F}_{1}$ and $\vec{F}_{2}$ act on a 4.00 kg disk that slides over frictionless ice, on which an $x y$ coordinate system is laid out. Force $\vec{F}_{1}$ is in the positive $x$ direction and has a magnitude of 7.70 N . Force $\vec{F}_{2}$ has a magnitude of 9.10 N . The figure here gives the $x$ component $v_{x}$ of the velocity of the disk as a function of time $t$ during the sliding. What is the angle between the constant directions of forces $\vec{F}_{1}$ and $\vec{F}_{2}$ ? Number i Unit

Solution

Solution Steps

Step 1: Determine the acceleration from the velocity-time graph

From the graph, we can see that the velocity $ v_x $ changes linearly with time $ t $. The slope of the line gives the acceleration $ a_x $.

\[ a_x = \frac{\Delta v_x}{\Delta t} = \frac{4 \, \text{m/s} - (-4 \, \text{m/s})}{3 \, \text{s} - 0 \, \text{s}} = \frac{8 \, \text{m/s}}{3 \, \text{s}} = \frac{8}{3} \, \text{m/s}^2 \]

Step 2: Calculate the net force in the x-direction

Using Newton's second law, $ F = ma $, we can find the net force $ F_x $ acting on the disk in the x-direction.

\[ F_x = m \cdot a_x = 4.00 \, \text{kg} \cdot \frac{8}{3} \, \text{m/s}^2 = \frac{32}{3} \, \text{N} \]

Step 3: Resolve the forces into components and find the angle

Given that $ \vec{F}_1 $ is in the positive x-direction with a magnitude of 7.70 N, and $ \vec{F}_2 $ has a magnitude of 9.10 N, we need to find the angle $ \theta $ between $ \vec{F}_1 $ and $ \vec{F}_2 $.

The net force in the x-direction is the sum of the x-components of $ \vec{F}_1 $ and $ \vec{F}_2 $:

\[ F_x = F_1 + F_2 \cos(\theta) \]

Substituting the known values:

\[ \frac{32}{3} \, \text{N} = 7.70 \, \text{N} + 9.10 \, \text{N} \cos(\theta) \]

Solving for $ \cos(\theta) $:

\[ \frac{32}{3} - 7.70 = 9.10 \cos(\theta) \]

\[ \frac{32}{3} - \frac{23.1}{3} = 9.10 \cos(\theta) \]

\[ \frac{8.9}{3} = 9.10 \cos(\theta) \]

\[ \cos(\theta) = \frac{8.9}{3 \cdot 9.10} \]

\[ \cos(\theta) = \frac{8.9}{27.3} \]

\[ \cos(\theta) \approx 0.326 \]

Finally, finding the angle $ \theta $: