Question

A box of books is initially at rest a distance \(D=0.540 \mathrm{~m}\) from the end of a wooden board. The coefficient of static friction between the box and the board is \(\mu_{s}=0.320\), and the coefficient of kinetic friction is \(\mu_{k}=0.250 .\) The angle of the board is increased slowly, until the box just begins to slide; then the board is held at this angle. Find the speed of the box as it reaches the end of the board. -4.55 A block of mass \(M_{1}=0.640 \mathrm{~kg}\) is initially at rest on a cart of mass \(M_{2}=0.320 \mathrm{~kg}\) with the cart initially at rest on a level air track. The coefficient of static friction between the block and the cart is \(\mu_{s}=0.620\), but there is essentially no friction between the air track and the cart. The cart is accelerated by a force of magnitude \(F\) parallel to the air track. Find the maximum value of \(F\) that allows the block to accelerate with the cart, without sliding on top of the cart.

Short answer

Answer: The box starts sliding down the board at an angle of θ = arctan(μs). The final velocity of the box when it reaches the end of the board can be found using the equation \(v_f = \sqrt{2 \cdot a \cdot D}\), where a is the acceleration of the box which can be calculated using Steps 2 and 3.

Step-by-step solution

Determine the angle of inclination when the box starts to slide

The box starts sliding when the maximum static friction force is equal to the component of the box's gravitational force parallel to the inclined plane. To find the angle, we use the formula for static friction force and gravitational force parallel component: \(\mu_s \cdot F_N = \mu_s \cdot m \cdot g \cdot \cos(\theta) = m \cdot g \cdot \sin(\theta)\) Here, \(\mu_s\) is the coefficient of static friction, \(F_N\) is the normal force (equal to \(m \cdot g \cdot \cos(\theta)\) for the inclined plane), \(m\) is the mass of the box, \(g\) is the gravitational acceleration, and \(\theta\) is the angle of the inclined plane. Solving for the angle, we get: \(\tan(\theta) = \frac{\mu_s}{\cos(\theta)}\) \(\theta = \arctan(\mu_s)\)

Find the acceleration of the box

Now, we'll use the coefficient of kinetic friction, \(\mu_k\), to find the acceleration of the box down the incline. The kinetic friction force acting on the box is: \(F_f = \mu_k \cdot F_N = \mu_k \cdot m \cdot g \cdot \cos(\theta)\) The net force acting on the box parallel to the incline is: \(F_{net} = m \cdot g \cdot \sin(\theta) - F_f\) Using Newton's second law of motion, we can find the acceleration of the box as follows: \(F_{net} = m \cdot a\) \(a = \frac{F_{net}}{m}\)

Calculate the final velocity of the box

We can now use the kinematic equation to find the final velocity, \(v_f\), of the box when it reaches the end of the board at distance \(D\): \(v_f^2 = v_i^2 + 2 \cdot a \cdot D\) The initial velocity of the box (\(v_i\)) is zero, so the equation becomes: \(v_f^2 = 2 \cdot a \cdot D\) \(v_f = \sqrt{2 \cdot a \cdot D}\) Now, substitute the given values of \(D\), \(\mu_s\), and \(\mu_k\) into the equations in Steps 1, 2, and 3 to find the final velocity of the box as it reaches the end of the board.

Key concepts

Static Friction

Static friction is an essential concept in physics, especially when objects are at rest relative to each other. This type of friction prevents the motion of an object when it is subjected to an external force. The force of static friction arises due to the interactions between the surfaces in contact and can vary depending on the types of materials and how much they are pressed together.

When you're solving a problem involving static friction, such as a box resting on an inclined plane, it's important to remember that this frictional force will adjust to match the applied force up to its maximum limit. This limit is defined by the coefficient of static friction, represented as \(\mu_s\), and the normal force, which is the perpendicular contact force between the two surfaces.

For instance, in our box example, only when the angle of the incline increases enough that the component of gravitational force parallel to the plane exceeds the maximum static friction, does the box begin to slide. The angle at which this occurs is critical because it marks the transition between no motion and the start of movement.

Kinetic Friction

Once the box starts moving, kinetic friction takes over. This is the frictional force opposing the motion of an object sliding over a surface. While static friction acts to prevent motion, kinetic friction works to reduce the motion that's already happening.

Unlike static friction, kinetic friction remains constant regardless of the speed the object is moving. It's determined by the coefficient of kinetic friction, \(\mu_k\), and the normal force. In the context of our sliding box, once it begins to descend the incline, \(\mu_k\) helps us calculate the force of kinetic friction that's slowing it down. This frictional force ultimately affects the box's acceleration as it travels down the plane.

Remember, kinetic friction is usually less than static friction, which is why an object in motion will continue to move until other forces (like kinetic friction or another external force) act upon it to slow it down and eventually stop it. In problem-solving, calculating kinetic friction is vital for finding out an object's acceleration and velocity when in motion, as outlined in the steps provided in the solution for the sliding box.

Inclined Plane

An inclined plane, often referred to as a ramp, is a flat surface tilted at an angle to the horizontal. This simple machine aims to facilitate the lifting of heavy objects by distributing the weight over a longer distance, reducing the necessary lifting force.

The study of objects on inclined planes is fundamental in physics since it involves analyzing the forces acting on the object, such as gravity, normal force, and friction. Gravitational force can be broken down into two components when dealing with an inclined plane: the force acting perpendicular to the plane (normal force) and the force acting parallel to the plane (which causes the object to slide down).

When solving problems with inclined planes, you'll typically decompose the forces, apply Newton's second law, and use the geometry of the plane (such as its angle of inclination) to determine various outcomes, like the acceleration of an object down the plane and its final speed. An important takeaway is that the steeper the angle of the inclined plane, the greater the parallel component of the gravitational force, and, therefore, the greater likelihood the object will overcome static friction and begin to move.