To solve this problem, we need to analyze the forces acting on the box as it moves along the ramp defined by the equation \( y = 5 - \frac{1}{10} x^2 \). We will determine the normal force acting on the box and its acceleration at the point where \( x = 5 \) meters.
Understanding the Ramp's Shape
The equation of the ramp, \( y = 5 - \frac{1}{10} x^2 \), describes a downward-opening parabola. At \( x = 5 \), we can find the height \( y \) of the ramp:
- Substituting \( x = 5 \) into the equation gives:
- \( y = 5 - \frac{1}{10} (5^2) = 5 - \frac{25}{10} = 5 - 2.5 = 2.5 \) meters.
So, at \( x = 5 \), the box is at a height of 2.5 meters on the ramp.
Calculating the Normal Force
The normal force is the force exerted by a surface that supports the weight of an object resting on it. It acts perpendicular to the surface. To find the normal force, we first need to determine the angle of the ramp at \( x = 5 \).
Finding the Slope of the Ramp
The slope of the ramp can be found by taking the derivative of the equation:
- \( \frac{dy}{dx} = -\frac{1}{5} x \)
- At \( x = 5 \), the slope is \( \frac{dy}{dx} = -\frac{1}{5} (5) = -1 \).
This slope indicates that the ramp is inclined at an angle \( \theta \) where \( \tan(\theta) = -1 \), which means \( \theta = 45^\circ \).
Weight and Normal Force Calculation
The weight of the box can be calculated using the formula:
- Weight \( W = mg \), where \( m = 30 \) kg and \( g \approx 9.81 \) m/s².
- Thus, \( W = 30 \times 9.81 = 294.3 \) N.
The normal force \( N \) can be calculated using the formula:
- \( N = W \cos(\theta) \)
- Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \), we have:
- \( N = 294.3 \times \frac{1}{\sqrt{2}} \approx 208.2 \) N.
Finding the Acceleration of the Box
Next, we need to find the acceleration of the box at \( x = 5 \). The net force acting on the box along the ramp can be determined by considering the component of gravitational force acting down the ramp.
Calculating the Gravitational Force Component
The component of the weight acting down the ramp is given by:
- \( F_{\text{gravity}} = W \sin(\theta) \)
- Using \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \), we find:
- \( F_{\text{gravity}} = 294.3 \times \frac{1}{\sqrt{2}} \approx 208.2 \) N.
Applying Newton's Second Law
According to Newton's second law, the net force is equal to mass times acceleration:
- \( F_{\text{net}} = ma \)
- Thus, \( a = \frac{F_{\text{net}}}{m} \).
Since the only force acting down the ramp is the gravitational component, we have:
- \( a = \frac{F_{\text{gravity}}}{m} = \frac{208.2}{30} \approx 6.94 \) m/s².
Summary of Results
At the point where \( x = 5 \) meters on the ramp:
- The normal force acting on the box is approximately \( 208.2 \) N.
- The acceleration of the box is approximately \( 6.94 \) m/s² down the ramp.
This analysis shows how the forces interact on an inclined surface and how to apply basic principles of physics to solve real-world problems. If you have any further questions or need clarification on any step, feel free to ask!
