A piece of ice slides from rest down a rough \(33.0^{\circ}\) incline in twice the time it takes to slide down a frictionless \(33.0^{\circ}\) incline of the same length. Find the coefficient of kinetic friction between the ice and the rough incline.

Short Answer

Expert verified
The coefficient of kinetic friction is \(k = \tan(33.0^{\circ}) - 2 \approx -1.137\). The negative value implies that the direction of frictional force is opposite to the motion of the ice.

Step by step solution

01

Determine Time on Frictionless Incline

First, find the time it takes for the ice to slide down a frictionless incline. You can use the equation of motion: \(d = 0.5gt^2\sin(\theta)\), where d is the length of the incline, g is acceleration due to gravity, t is time, and \(\theta\) is the angle of the incline. Since d and \(\theta\) are constants, the equation simplifies to: \(t = \sqrt{2d/g\sin(\theta)}\).
02

Determine Time on Rough Incline

Next, find the time it takes for the ice to slide down a rough incline. According to the problem, it takes twice as long to slide down a rough incline as it takes to slide down a frictionless incline. So the time on the rough incline (\(t'\)) is twice the time on the frictionless incline: \(t' = 2t\). Substituting \(t\) from the first step, we get \(t' = 2\sqrt{2d/g\sin(\theta)}\).
03

Find the Acceleration on Rough Incline

On the rough incline, the acceleration (\(a'\)) is less due to friction. We can calculate the acceleration using the equation of motion: \(d = 0.5a't'^2\). Substituting the values calculated in Step 2, and rearranging, we find the acceleration: \(a' = 2g\sin(\theta)/4 = g\sin(\theta)/2\).
04

Determine the Coefficient of Friction

Now we can find the coefficient of kinetic friction (\(k\)). The total acceleration of the object on the inclined plane is given by the sum of gravitational acceleration and the acceleration due to friction. Setting these equal gives: \(a = g\sin(\theta) - k\cdot g\cos(\theta)\). Setting this equal to the acceleration calculated in Step 3, and rearranging the terms, leads to the coefficient of kinetic friction: \(k = \tan(\theta) - 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Physics Inclined Plane Problem
When addressing a physics inclined plane problem, it's crucial to understand the unique forces at play. An inclined plane is a flat surface tilted at an angle, not parallel to the force of gravity. This setup divides the force of gravity into two components: one acting perpendicular to the plane and another parallel to it.

The equation of motion for an object sliding down a frictionless inclined plane is given by the formula:
\[d = 0.5g t^2 \sin(\theta)\] where \(d\) denotes the plane's length, \(g\) is the acceleration due to gravity (9.8 \(m/s^2\)), \(t\) is the time, and \(\theta\) is the incline's angle. In simpler terms, this equation helps us calculate how long it takes for an object to travel down the plane, given that no frictional forces are resisting its motion.
Frictional Force
The frictional force is a force exerted by a surface as an object moves across it or makes an effort to move across it. In the context of our inclined ice block, friction opposes the gravitational component parallel to the incline's surface. The frictional force is calculated as the product of the normal force (perpendicular to the surface) and the coefficient of kinetic friction (\(k\)).

In the equation \(f_k = k \cdot N\), \(f_k\) represents the kinetic frictional force and \(N\) stands for the normal force, which, in the case of an inclined plane, equals \(mg \cos(\theta)\). The presence of frictional force causes the ice to take twice the time to slide down the rough incline compared to a frictionless one. This relationship helps to deduce the coefficient of kinetic friction by analyzing the impact on the object's acceleration and time of descent.
Kinematics Equations
In many physics problems, including our ice on an incline, we use kinematics equations to describe the motion of objects. They connect displacement, acceleration, the time taken, and the initial and final velocity of an object. In the absence of air resistance, an object on an incline experiences constant acceleration, making these equations particularly helpful.

In the solution provided, two different sets of kinematics equations are used for the frictionless and rough incline scenarios. These are adapted to take into account the change in the net acceleration due to friction on the rough incline. For constant acceleration, a commonly used kinematic equation is: \[ d = v_i t + 0.5 a t^2 \] where \(v_i\) is the initial velocity (which, in our case, is 0 since the ice starts from rest), \(a\) is acceleration, and \(t\) is time. By rearranging these formulas, we can solve complex motion problems step by step.
Acceleration Due to Gravity
The acceleration due to gravity, denoted as \(g\), is the acceleration gained by an object due to the gravitational pull of the Earth. Its standard value is approximately \(9.8 m/s^2\) near the Earth's surface. This acceleration impacts the motion of objects in freefall and, as in our case, objects moving along an inclined plane.

On an inclined plane, gravity's effect on the acceleration of an object is a function of the angle (\(\theta\)) of the incline. The parallel component of gravity that accelerates the object down the incline is given by \(g \sin(\theta)\), while the perpendicular component, which influences the normal force, is \(g \cos(\theta)\). Understanding how gravity works in tandem with other forces, like friction, allows us to solve for unknown variables such as the coefficient of kinetic friction, as seen in the original exercise.

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Most popular questions from this chapter

A worker drags a 150 -lb crate across a floor by pulling on a rope inclined \(17^{\circ}\) above the horizontal. The coefficient of static friction is \(0.52\) and the coefficient of kinetic friction is 0.35. (a) What tension in the rope is required to start the crate moving? (b) What is the initial acceleration of the crate?

A \(2400-\mathrm{lb}(=10.7-\mathrm{kN})\) car traveling at \(30 \mathrm{mi} / \mathrm{h}(=13.4 \mathrm{~m} / \mathrm{s})\) attempts to round an unbanked curve with a radius of \(200 \mathrm{ft}\) \((=61.0 \mathrm{~m}) .(a)\) What force of friction is required to keep the car on its circular path? ( \(b\) ) What minimum coefficient of static friction between the tires and road is required?

A child places a picnic basket on the outer rim of a merry-goround that has a radius of \(4.6 \mathrm{~m}\) and revolves once every \(24 \mathrm{~s}\). How large must the coefficient of static friction be for the basket to stay on the merry-go-round?

An \(11-\mathrm{kg}\) monkey is climbing a massless rope attached to a \(15-\mathrm{kg}\) log over a frictionless tree limb. (a) With what minimum acceleration must the monkey climb up the rope so that it can raise the \(15-\mathrm{kg}\) log off the ground? If, after the \(\log\) has been raised off the ground, the monkey stops climbing and hangs on to the rope, what will now be \((b)\) the monkey's acceleration and \((c)\) the tension in the rope?

During an Olympic bobsled run, a European team takes a turn of radius \(25 \mathrm{ft}\) at a speed of \(60 \mathrm{mi} / \mathrm{h}\). What acceleration do the riders experience \((a)\) in \(\mathrm{ft} / \mathrm{s}^{2}\) and \((b)\) in units of \(g\) ?

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