Video: Kinematics of an Object Sliding Down a Slope against Friction

A box of mass 12.4 kg slides from rest downward along a 1.58-m-length ramp that is inclined at 23.3Β° below the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.0346. What is the magnitude of the box’s acceleration? What is the speed of the box at the bottom of the ramp?

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Video Transcript

A box of mass 12.4 kilograms slides from rest downward along a 1.58-meter-length ramp that is inclined at 23.3 degrees below the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.0346. What is the magnitude of the box’s acceleration? What is the speed of the box at the bottom of the ramp?

We can call the box’s acceleration π‘Ž and the box’s speed at the bottom of the ramp 𝑣 sub 𝑓. Let’s start on our solution by drawing a diagram of the moving box. In this example, we have a box of mass π‘š β€” π‘š is 12.4 kilograms β€” on a ramp of length 𝑙, where 𝑙 is 1.58 meters. The ramp is inclined at an angle we’ve called πœƒ, where πœƒ is 23.3 degrees above the horizontal. Under the influence of gravity, we expect the box to slide down the ramp. and we want to solve for its acceleration, π‘Ž, when it does so. We also wanna find out how fast the box is moving when it gets to the bottom of the ramp.

Into this scenario, we set up an π‘₯- and a 𝑦-coordinate axis so that positive motion in the π‘₯-direction is down the incline and positive motion in the 𝑦-direction is perpendicular to the plane. Next, let’s draw a free body diagram of the mass π‘š, that is, the forces that act on it. We know there’s a gravitational force acting on π‘š straight down of magnitude π‘š times 𝑔, where 𝑔, the acceleration due to gravity, is 9.8 meters per second squared. There’s also a normal force acting on the mass to keep it from moving into the plane. And lastly, as the box slides down the plane, there’s a kinetic frictional force that resists its motion.

Looking at this free body diagram, we can divide up the weight force into its 𝑦- and π‘₯-components so that we can write out force balance equations in the 𝑦- and π‘₯-directions. And we know that, with these force components at a right angle to one another, a triangle is formed, where the upper angle is the angle πœƒ.

Recalling Newton’s second law of motion, that the net force on an object equals its acceleration times its mass, we can consider the forces in the π‘₯-direction in our scenario. Those are the kinetic friction force and the π‘₯-component of the weight force. Since we’ve decided that motion down the ramp is motion in the positive direction, our force balance equation for the π‘₯-forces is π‘šπ‘” sin πœƒ minus the kinetic friction force is equal to the mass of the object times its acceleration.

The kinetic friction force, 𝑓 sub π‘˜, is equal to the coefficient of kinetic friction, πœ‡ sub π‘˜, multiplied by the normal force acting on an object. This means we can replace 𝑓 sub π‘˜ with πœ‡ sub π‘˜ 𝑓 sub 𝑛 in our π‘₯-force balance equation. And we remember that we’re given πœ‡ sub π‘˜ in the problem statement.

Looking at this force balance equation, we see that if we divide both sides by the mass π‘š, we see we have an expression for the acceleration π‘Ž we want to solve for. The one challenge is, we don’t yet know the normal force, 𝑓 sub 𝑛, that acts on our mass. To find that out, let’s look at the forces in the 𝑦-direction on our mass.

Applying Newton’s second law to forces in this direction, we write that the normal force minus π‘šπ‘” cos πœƒ, the 𝑦-component of the weight force, is equal to the mass of the box times its acceleration in the 𝑦-direction. But because the box doesn’t leave the surface of the plane, this acceleration is zero. And our equation simplifies to the normal force is equal to π‘šπ‘” times the cos of πœƒ. This is perfect because now we can take this term for 𝑓 sub 𝑛 and substitute it into our equation in the π‘₯-direction.

With this expanded expression for the acceleration π‘Ž, we see that the box’s mass π‘š appears in all the terms in the numerator and the denominator. So it cancels out. Factoring out the acceleration due to gravity, π‘Ž is equal to 𝑔 times the quantity sin πœƒ minus πœ‡ sub π‘˜ cos πœƒ. Substituting in for 𝑔, πœƒ, and πœ‡ sub π‘˜, when we calculate π‘Ž, to three significant figures, we find it’s 3.56 meters per second squared. That’s the acceleration of the box as it descends the ramp.

Next, we wanna solve for the speed the box has when it reaches the bottom of the 1.58-meter-long ramp. To figure this out, we know that the acceleration the box undergoes during its journey is constant. This means that the kinematic equations of motion apply to the motion of this box. As we look over these four equations of motion, we seek one that lets us solve for final velocity, 𝑣 sub 𝑓, in terms of information we know.

The second kinematic equation written matches our conditions well. Rewriting it in terms of our variables, the final speed of the box squared is equal to the box’s initial speed, which is zero because the box starts from rest, plus two times the box’s acceleration times the length of the ramp, 𝑙. If we take the square root of both sides of this equation, we find that 𝑣 sub 𝑓 is equal to the square root of two times π‘Ž times 𝑙.

We know π‘Ž from part one and 𝑙 is given information. So we’re ready to plug those in to our expression. When we do, when enter these values on our calculator, we find that 𝑣 sub 𝑓 is 3.36 meters per second. That’s the speed of the box when it reaches the bottom of the ramp.

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