🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning

Like

Report

Numerade Educator

Like

Report

Problem 93 Hard Difficulty

ssm A penguin slides at a constant velocity of 1.4 $\mathrm{m} / \mathrm{s}$ down an icy incline. The incline slopes above the horizontal at an angle of $6.9^{\circ} .$ At the bottom of the incline, the penguin slides onto a horizontal patch of ice. The coefficient of kinetic friction between the penguin and the ice is the same for the incline as for the horizontal patch. How much time is required for the penguin to slide to a halt after entering the horizontal patch of ice?

Answer

1.2 $\mathrm{s}$

Discussion

You must be signed in to discuss.

Video Transcript

In order to solve this question, we have to discover what is the acceleration of the penguin in the second step off his business. In order to do that, we have to apply Newton's second law the second situation as my reference frame. For the second situation, I would choose thes one, a vertical axis and a horizontal axis. I would call the vertical axis. Why, under horizontal Axis X, then applying Newton's second law situation to on the horizontal access we got. The following the net force on the horizontal direction is because of the mass off the Penguin Times, its acceleration on the result without action. The net force in the horizontal direction is composed in the second situation on Lee by the frictional force, which points to the left. So minus frictional force is because the mass off the Penguin times he's acceleration before the acceleration off the penguin is the question minus frictional force divided by his mass. But in the second situation, the penguin is already moving, so the frictional force is given by the kinetic frictional coefficient times the normal force, then his acceleration next direction is a question minus the kinetic frictional coefficient times the normal force divided by his mass. But on the second situation, it's very easy to see that the normal force is exactly compensated by the weight force before the normal force is because of the weight force in these second situation, Then using that normal force is it was the weight force, which is because of the mass times acceleration of gravity. We concluded that his acceleration that's direction is he goes to minus the kinetic frictional coefficient times his mass times, the acceleration of gravity divided by his mess. Then we can simplify the masses here to get an expression for his acceleration. So his acceleration is the question minus the kinetic frictional coefficient times the acceleration of gravity. In order to use this equation, we have to discover what is the kinetic friction coefficient. And for that, we have to look at Situation number one, where he's sliding down a slope in situation number one. I would choose another reference frame on peak these one, and you call these Axis X and these other access. Why, then, applying Newton's second law, that's question number one to the horizontal axis so that we can get a relation with the frictional force, we get the following the net force. The X direction is equal to the mass off the Penguin times his acceleration in that direction. But his is lighting with constant velocity instead, number one. Therefore his acceleration is close to zero. Then the net force is equal to zero. But the net force in the X direction is given by two forces the frictional force and the X component off the weight force, which can be decomposed as follows. There is a X component w x on a Y component, though you weigh so the net force can be returning us the weight component x minus. The frictional force on this is the course to zero. Then the wait for us component acts is echoes the frictional force again. The frictional force in the situation is the kinetic frictional force. So it's given by the kinetic frictional coefficient times their normal force. Therefore, the weight component acts easy. Of course. True, the kinetic frictional coefficient times the normal force. No, we can start. So see from this figure that if we apply Newton's second Law on the Y axis, we will end up in the conclusion that the normal force is equal to the weight forced. Why component? Because he isn't moving in these axis. So the acceleration is close to zero. And at the same time, the net force is composed by two forces the normal force and the weight component way. So we conclude that the normal force is equal to the weight component. Why the reform? The weight component acts is equals tomb UK times the weight component. Why? Because the normal force is the cause. That's no way to conform it. Why? Damn the kinetic frictional coefficient is the coastal w acts divided by the value. Why? And what is this for that we have to take a look at the triangle that is formed by the weight force on its components. So here we have the weight force year. We have the weight force. Why component? And here we have the weight forced X component. Then we have to discover an angle in this triangle. For that, we can extend the weight forced down here so that it forms a 90 degree angle. So we have a triangle like this. This triangle has a 90 degree angle 6.9 degrees angle and another angle in order for the strangle toe have 180 degrees as a result, off the sun off its intern angles. This I know here must be 83 0.1 degrees at the same time. This angle is ouster 90 degrees, then that in turn angle that we have here it must be such that when you add this angle and a 3.1 degrees, you get 90 degrees. Therefore, this angle it's also 6.9 degrees. Then the conclusion is that the angle between the Y component on the weight force is 6.9 degrees. So this angle 6.9 degrees, this is the weight force. Then we can calculate to the tangent off 6.19 reasons follows. It is equals to the opposite site w X, divided by the interests inside the were you right? Then we can use these result in his expression to get that the kinetic frictional coefficient is equals to the tangent off 6.9 degrees. Then now that we know, everybody is the kinetic frictional coefficient, we can go back and calculate what is the acceleration off the penguin in the second step off the situation. So he's acceleration is it goes to minus the tangent off 6.9 degrees times the acceleration of gravity, which is approximately 9.8 meters per second squared this give us an acceleration off approximately minus 1.186 meters per second squared. This is the acceleration off the dangling. Then we can now calculate How much time does it take for the pain wind come toe help after entering the second phase off the movement In that second phase of the movement, the velocity as a function of time is given by his initial velocity, plus his acceleration times fainter off time. We know that his final velocity will be close to zero. We also know that his initial velocity is 1.4 meters per second. Then the acceleration is minus 1.186 and then we multiply it by the time it takes to come toe health. So we have to solve this equation for tea. We can do this by sending this term to the other side. So 1.1 it six times t is it goes to 1.4 and then sending this term to the other side, so T is equals to 1.4, divided by 1.186 which results in the time off approximately one 0.2 seconds. And this is the time it takes for the Penguin to stop moving after reaching the second things off the movement.