conceptual activity
1. explore the simulation on your own for several minutes.
2. set angle of inclination to 30°, coefficient of kinetic friction to 0, and coefficient of static friction to 0.00.
3. begin with mass of block = 0.50 kg. select go. record the net force and the acceleration in table 1.
4. repeat the process while increasing the mass by 0.50 kg increments. fill in table 1 as you collect data.
observations and analysis
table 1 (θ = 30°, μs = 0.00; μk = 0.00)

Question

Accepted Answer

To solve for the net force and acceleration in this inclined - plane problem (with no friction, $\mu_s = 0.00$ and $\mu_k=0$), we use the following physics concepts:

### Step 1: Recall the formula for the component of gravitational force along the incline
The force due to gravity acting on an object of mass $m$ is $F_g = mg$, where $g = 9.8\space m/s^2$ (acceleration due to gravity). The component of this force along the incline (when the angle of inclination is $\theta$) is given by $F_{net}=mg\sin\theta$ (since there is no friction, the net force along the incline is just this component).

### Step 2: Recall the formula for acceleration from Newton's second law
Newton's second law states that $F = ma$. Since $F_{net}=mg\sin\theta$, we can solve for acceleration $a$ by dividing both sides of the equation $F_{net}=ma$ by $m$. We get $a = g\sin\theta$.

Now let's calculate the net force and acceleration for each mass:

#### For $m = 0.50\space kg$:
- **Net Force ($F_{net}$)**:
Using $F_{net}=mg\sin\theta$, with $m = 0.50\space kg$, $g = 9.8\space m/s^2$ and $\theta = 30^{\circ}$ (and $\sin30^{\circ}=\frac{1}{2}=0.5$)
$F_{net}=0.50\times9.8\times\sin30^{\circ}$
$F_{net}=0.50\times9.8\times0.5$
$F_{net}=2.45\space N$
- **Acceleration ($a$)**:
Using $a = g\sin\theta$
$a=9.8\times\sin30^{\circ}$
$a = 9.8\times0.5=4.9\space m/s^2$

#### For $m = 1.00\space kg$:
- **Net Force ($F_{net}$)**:
$F_{net}=1.00\times9.8\times\sin30^{\circ}$
$F_{net}=1.00\times9.8\times0.5 = 4.9\space N$
- **Acceleration ($a$)**:
$a = 9.8\times\sin30^{\circ}=4.9\space m/s^2$ (acceleration is independent of mass in the absence of friction)

#### For $m = 1.50\space kg$:
- **Net Force ($F_{net}$)**:
$F_{net}=1.50\times9.8\times\sin30^{\circ}$
$F_{net}=1.50\times9.8\times0.5=7.35\space N$
- **Acceleration ($a$)**:
$a = 9.8\times\sin30^{\circ}=4.9\space m/s^2$

#### For $m = 2.00\space kg$:
- **Net Force ($F_{net}$)**:
$F_{net}=2.00\times9.8\times\sin30^{\circ}$
$F_{net}=2.00\times9.8\times0.5 = 9.8\space N$
- **Acceleration ($a$)**:
$a = 9.8\times\sin30^{\circ}=4.9\space m/s^2$

#### For $m = 2.50\space kg$:
- **Net Force ($F_{net}$)**:
$F_{net}=2.50\times9.8\times\sin30^{\circ}$
$F_{net}=2.50\times9.8\times0.5=12.25\space N$
- **Acceleration ($a$)**:
$a = 9.8\times\sin30^{\circ}=4.9\space m/s^2$

#### For $m = 3.00\space kg$:
- **Net Force ($F_{net}$)**:
$F_{net}=3.00\times9.8\times\sin30^{\circ}$
$F_{net}=3.00\times9.8\times0.5 = 14.7\space N$
- **Acceleration ($a$)**:
$a = 9.8\times\sin30^{\circ}=4.9\space m/s^2$

We can tabulate the results as follows:

| Mass (kg) | Net Force (N) | Acceleration ($m/s^2$) |
|---|---|---|
| 0.50 | 2.45 | 4.9 |
| 1.00 | 4.9 | 4.9 |
| 1.50 | 7.35 | 4.9 |
| 2.00 | 9.8 | 4.9 |
| 2.50 | 12.25 | 4.9 |
| 3.00 | 14.7 | 4.9 |

### Explanation of the pattern
- The net force is directly proportional to the mass ($F_{net}\propto m$) because $F_{net}=mg\sin\theta$ and $g\sin\theta$ is a constant in this case (since $\theta$ and $g$ are fixed).
- The acceleration is constant ($a = g\sin\theta$) and independent of the mass. This is because when we apply $F = ma$ and $F = mg\sin\theta$, the mass $m$ cancels out ($a=\frac{F}{m}=\frac{mg\sin\theta}{m}=g\sin\theta$), which is a demonstration of Galileo's principle that in the absence of air resistance (or in this case, friction), all objects fall (or accelerate down an incline) at the same rate regardless of their mass.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step 1: Recall the formula for the component of gravitational force along the incline

Step 2: Recall the formula for acceleration from Newton's second law

For \(m = 0.50\space kg\):

For \(m = 1.00\space kg\):

For \(m = 1.50\space kg\):

For \(m = 2.00\space kg\):

For \(m = 2.50\space kg\):

For \(m = 3.00\space kg\):

Answer: