Problem 36 - University Physics with Modern Physics

Question

A 25.0 -kg box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is \(0.25,\) and the coefficient of static friction is \(0.35 .\) (a) As the angle \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0 \(\mathrm{m}\) along the loading ramp?

Step-by-Step Solution

Answer

(a) \(19.29^\circ\); (b) \(1.18 \text{ m/s}^2\); (c) \(3.44 \text{ m/s}\).

Step 1: Calculate Critical Angle for Slipping

To determine the angle \( \alpha \) at which the box begins to slip, we use the static friction equation. The box will slip when the gravitational component along the ramp equals the maximum static friction force. This is given by:\[\tan(\alpha) = \frac{\mu_s \cdot g \cdot m \cdot \cos(\alpha)}{g \cdot m \cdot \sin(\alpha)} = \mu_s\]where \( \mu_s = 0.35 \). Solving for \( \alpha \), we get:\[\alpha = \arctan(\mu_s) = \arctan(0.35) \approx 19.29^\circ\]

Step 2: Calculate Normal Force and Friction Force

Once the box starts moving, the normal force \( N \) and kinetic friction force \( f_k \) are important. On an inclined plane:\[N = m \cdot g \cdot \cos(\alpha) \f_k = \mu_k \cdot N = \mu_k \cdot m \cdot g \cdot \cos(\alpha)\]Substituting \( \alpha = 19.29^\circ \), \( \mu_k = 0.25 \), \( m = 25.0 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \),we find\( N \approx 230.6 \text{ N} \) and \( f_k \approx 57.6 \text{ N} \).

Step 3: Calculate Acceleration of the Box

The net force along the ramp determines the box's acceleration. Use Newton's second law:\[F_{net} = m \cdot g \cdot \sin(\alpha) - f_k\]Thus, acceleration \( a \) is:\[a = \frac{F_{net}}{m} = g \cdot \sin(\alpha) - \mu_k \cdot g \cdot \cos(\alpha)\]Substituting the given values:\[a \approx 9.8 \cdot \sin(19.29^\circ) - 0.25 \cdot 9.8 \cdot \cos(19.29^\circ) \approx 1.18 \text{ m/s}^2\]

Step 4: Calculate Velocity after 5.0 m Slide

Use the equation of motion to find the velocity after sliding 5.0 m:\[v^2 = u^2 + 2a s\]where \( u = 0 \text{ m/s} \) (initially at rest), \( a = 1.18 \text{ m/s}^2 \), \( s = 5.0 \text{ m} \).Thus,\[v^2 = 0 + 2 \cdot 1.18 \cdot 5 = 11.8\]\[v = \sqrt{11.8} \approx 3.44 \text{ m/s}\]

Key Concepts

Static Friction

Static friction is the force that keeps an object at rest when it is placed on a surface. It acts in the opposite direction to the applied force. For the 25.0-kg box on the loading ramp, static friction is what initially prevents the box from sliding down. The maximum force of static friction can be calculated using the equation:

• \[f_s = \mu_s \cdot N\]

where \(\mu_s\) is the coefficient of static friction, and \(N\) is the normal force, which in this context is equal to the gravitational pull perpendicular to the ramp: \(m \cdot g \cdot \cos(\alpha)\).
By determining the critical angle \(\alpha_c\) such that the force pulling the box down the ramp overcomes the maximum static friction, we find when slipping begins.In the exercise, we solved for this angle using:

• \[\tan(\alpha_c) = \mu_s\]

For our given \(\mu_s = 0.35\), this results in \(\alpha = \arctan(0.35)\).

This means the box begins to slide at approximately \(19.29^\circ\).
Static friction is all about resisting the initial move until the force down the slope exceeds this calculated maximum friction force.

Kinetic Friction

Kinetic friction acts on objects that are moving. Unlike static friction, kinetic friction is usually less intense, allowing the object to continue moving once it has started. The force of kinetic friction can be calculated using the equation:

• \[f_k = \mu_k \cdot N\]

where \(\mu_k\) is the coefficient of kinetic friction, and \(N\) is the normal force.
In our scenario, once the box starts moving down the incline, kinetic friction takes over, opposing the movement.
The box on the ramp will experience a kinetic friction calculated as follows:Given \(\mu_k = 0.25\), we use

• \[f_k = 0.25 \cdot 25.0 \text{ kg} \cdot 9.8 \text{ m/s}^2 \cdot \cos(19.29^\circ)\]

This results in a kinetic friction force \( f_k \approx 57.6 \text{ N} \).
So, after the box starts slipping, kinetic friction will moderate the speed at which the box accelerates down the ramp. This is a key part of understanding motion over slopes in physics.

Newton's Second Law

Newton's Second Law is one of the most important principles for understanding motion. It states that the net force \(F_{net}\) acting on an object is equal to the mass \(m\) of that object multiplied by its acceleration \(a\):

• \[F_{net} = m \cdot a\]

In the inclined plane scenario, the net force driving the box down the ramp is the difference between the gravitational force component along the incline and the opposing kinetic friction.Mathematically, it is expressed as:

• \[F_{net} = m \cdot g \cdot \sin(\alpha) - f_k\]

Substitute \(m = 25.0 \text{ kg},\ \alpha = 19.29^\circ,\ \text{and}\ f_k = 57.6 \text{ N}\) into the equation.
Using this principle helps us determine the box's acceleration on the ramp by simply rearranging the formula to solve for \(a\):

• \[a = \frac{F_{net}}{m} = g \cdot \sin(19.29^\circ) - \mu_k \cdot g \cdot \cos(19.29^\circ)\]

Initially, you find that the calculation results in an approximate acceleration of \(1.18 \text{ m/s}^2\).
Understanding Newton's Second Law makes it easier to predict how objects will behave under the influence of different forces.

Acceleration Calculation

To determine how quickly something changes its velocity over time, we compute acceleration. In this problem, once the box starts moving, the acceleration can be found using the net force impact derived from the previous concepts. The acceleration \(a\) was calculated as:

• \[a = g \cdot \sin(19.29^\circ) - 0.25 \cdot g \cdot \cos(19.29^\circ)\]

After inserting the values, we found:

• \[a \approx 1.18 \text{ m/s}^2\]

To know how fast the box is moving after traveling 5.0 meters down the ramp, we can use the equation of motion:

• \[v^2 = u^2 + 2a s\]

where \(u = 0\) is the initial velocity, \(s = 5.0 \text{ m}\) is the distance traveled, and \(a\) is the calculated acceleration.
Plugging these values in results in:

• \[v^2 = 0 + 2 \cdot 1.18 \cdot 5\]

Calculating this, the box's velocity is approximately \(3.44 \text{ m/s}\) after 5 meters.
Mastering acceleration calculation allows us to understand how quickly an object's velocity will increase or decrease.