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

Like

Report

Numerade Educator

Like

Report

Problem 79 Hard Difficulty

ssm A student is skatcboarding down a ramp that is 6.0 $\mathrm{m}$ long and inclined at $18^{\circ}$ with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is 2.6 $\mathrm{m} / \mathrm{s}$ . Neglect friction and find the speed at the bottom of the ramp.

Answer

6.6 $\mathrm{m} / \mathrm{s}$

Discussion

You must be signed in to discuss.

Video Transcript

to solve this question, we have to determine what is the acceleration off the state order. In order to do that, we have to apply Newton's second law in the direction off amusement. You will move in this direction, then are you choose that everything pointing there is positive and everything pointing to the other side is negative. Then, applying Newton's second law on this direction, we get the following The net force in that direction is the goes to the mass off the skateboarder times. He is acceleration in this direction. Then the net force has only one component, which is this component off the weight force that he called his weight acts component. And then we also have this other component which is not off our interest now, which would be the Y component off the wait for us. Then, in a direction of movement, we have the weight force component acts pointing to the positive direction and these composers and that force. Then it is equal to the mass times acceleration off the skate border. But the axe component off the weight force can be related to the weight force by using at the composition off the weight factor. To do that, we have to take a look here in this triangle so it's directed will try and go and then we have to discover one off its intern angles. To do that, we note the following This is 18 degrees. Then we can stretch the weight forced down here so that these is 90 degrees them. If this is 18 degrees, these is 90 degrees. Then this angle must be 90 miners. 18 which is it Cause to 72 The grease, Then the relation between the weight force on the X component can be given by the co sign off 72 degrees because they co sign after 72 degrees is equals Truth the agile. It's inside too. The axe component off the weight force divided by the high pattern use, which is the weight force. Then the X component off the weight force is equal to the weight forced times. The co sign off 72 degrees. Then here we have the weight force times. The co sign off 72 degrees equals to the mass times acceleration off the skate Werder. But remember that the weight forces given by the mass off the skate border times. That's original gravity, so M times G times Nico sign off 72 degrees is he goes to the mass times acceleration off escape border. Then we can simplify the masses to get an acceleration that is equals g times the co sign off. 70 Truth. Remember that G is approximately 9.8 meters per second squared. So here we have 9.8 times the co sign off 72 degrees, and this is his acceleration. But we have to get her mind. What is the final velocity? So what is the velocity of the skate border at the end off that ramp? In order to do that, we now use Torricelli's equation, which says the following the final velocity squared is he goes to the initial velocity squared, plus true times, acceleration times the distance. Then the squared is it goes to true 0.6 squared plus true times. The acceleration, which is 9.8 times the co sign off 72 degrees times displacement, which is six meters. Then the velocity is the coast to the square. It of true 0.6 squared plus two times 9.8 times the co sign off 72 degrees times six. This gives a final velocity off approximately 6.6 meters per second.