A wooden plank is raised at one end to an angle of 30 degrees. A 2.0 kg box is placed on the incline 1.0 m from the lower end and given a slight tap to overcome static friction. The coefficient of a kinetic friction between the box and the plank is MUk = 0.20 . Find a) the rate of the acceleration of the box and b) the speed of the box at the bottom. Assume that the initial speed of the box is zero. Show ALL work in the space provided.

Question

UpStudy AI · Accepted Answer

**Step 1. Draw the Free-Body Diagram and Identify Forces**

For a box of mass $$ m = 2.0 \, \text{kg} $$ on an incline of angle $$ \theta = 30^\circ $$, the forces acting along the incline are:

- The component of gravity down the incline: 
  $$
  mg\sin\theta
  $$
- The kinetic friction force opposing the motion:
  $$
  f_k = \mu_k mg\cos\theta,
  $$
  where $$ \mu_k = 0.20 $$.

**Step 2. Write the Equation of Motion**

Taking the downward direction along the incline as positive, the net force acting on the box is:
$$
F_{\text{net}} = mg\sin\theta - \mu_k mg\cos\theta.
$$
By Newton's second law:
$$
ma = mg\sin\theta - \mu_k mg\cos\theta.
$$
Divide both sides by $$ m $$ to get the acceleration:
$$
a = g\left(\sin\theta - \mu_k\cos\theta\right).
$$

**Step 3. Substitute the Known Values**

Given:
$$
g = 9.8 \, \text{m/s}^2,\quad \sin(30^\circ) = 0.5,\quad \cos(30^\circ) \approx 0.866,
$$
and
$$
\mu_k = 0.20.
$$
Thus,
$$
a = 9.8\left(0.5 - 0.20 \times 0.866\right).
$$
Calculate the friction term:
$$
0.20 \times 0.866 \approx 0.1732,
$$
so:
$$
a = 9.8\left(0.5 - 0.1732\right) = 9.8 \times 0.3268 \approx 3.20 \, \text{m/s}^2.
$$

**Answer to Part (a):**  
The acceleration of the box down the incline is approximately $$ 3.20 \, \text{m/s}^2 $$.

**Step 4. Determine the Speed at the Bottom**

Using the kinematic equation for motion from rest:
$$
v^2 = v_0^2 + 2ad,
$$
where $$ v_0 = 0 $$ and $$ d = 1.0 \, \text{m} $$ (distance traveled along the plank), we have:
$$
v^2 = 0 + 2(3.20)(1.0) = 6.4.
$$
Taking the square root:
$$
v = \sqrt{6.4} \approx 2.53 \, \text{m/s}.
$$

**Answer to Part (b):**  
The speed of the box at the bottom of the incline is approximately $$ 2.53 \, \text{m/s} $$.
**Part (a):** The box accelerates down the incline at approximately $$ 3.20 \, \text{m/s}^2 $$. **Part (b):** The box reaches a speed of approximately $$ 2.53 \, \text{m/s} $$ at the bottom.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Physics Questions