questions-2022-refer-to-the-following-scenario-a-2-mathrmkg-ball-with-a-velocity-of-20-mathrmm-mat-3

Question

Questions 20–22 refer to the following scenario:
A 2 $\mathrm{kg}$ ball with a velocity of $-20 \mathrm{m} / \mathrm{s}$ collides with the wall and bounces back with a velocity of 10 $\mathrm{m} / \mathrm{s}$
How much work did the wall do on the ball?
(A) $-300 \mathrm{J}$
(B) $-400 \mathrm{J}$
(C) $-750 \mathrm{J}$
(D) $-1,000 \mathrm{J}$

Sachin Rao · Accepted Answer

this problem we have to find How much work did the ball do on the wall? So we know that net work them on the system is a call to change. In kinetic energy, that means K a final minus K initial. It&#x27;s okay. A finalist half and the final square, minus half m. We initially square so we can buy it. It is half into em. We finally square minus. We initially square. So this is the work done this half into Mars, which is to gauge e into the finalists. Tennis squared minus V initialism minus 20 square. Now we will solve this. This is a whole new mhm 100 minus 400. This is called a minus 300 jewels. This is that work done by the wall on the wall. If we check options, options, even we correct option. Option A. Yes, correct. Yes. Thank you.

Salamat Ali · Answer

The work done by the wall on the ball is given by the view is the call toe right off? Change of frantic energy. Not right off, Jane. Sorry. Changing Patrick Energy. The kind of energy change me. Harry is one of two turned the Maas final Velocity square a minus initial loss. T square, eso substituting. They&#x27;re aliens. One over to where mass we have is a two kg final lost in the heavy areas. There. 10. So square minus 20 square usually lost. So 20 square. This gives us the energy. So he worked on by the wall. On the ball is a minus 300 jewels, 300 jewels. So option is the right answer.

Salamat Ali · Answer

the average force experience by the ball when it hits the wall is force physical to rate of change of momentum. Dp derided by Delta T uh, where we have a DP From above question, we have 60 new two seconds divided by the time given. We have a 0.0. There are two seconds. So the writing these two numbers gives us the force over 30,000 Newton 30,000 You didn&#x27;t, So her option see is the right option.

Surjit Tewari · Answer

questionable 44 In the by bay, we have to find the kinetic energy at point A Hey, a is equal. Do, uh, times the mosque off the object into the velocity at the point. Esquire Substituting the values in it hard times Moss is zero point six giddy and we lost ease. No solving this. We will get the cannon. Take energy at point A as 1.2 George in that might be we have to find the velocity at point B and we know the value of kinetic energy so we can write 7.5. Jules, this is the kinetic energy at point B is equal to half times the mosque, which is 0.6 giddy into velocity at point B. It&#x27;s quiet solving this. We will get the velocity at one. B as five. Me that what a second? No. In that part of the question, we have to find that. But then we know that work then is equal. Duda change in Canada take energy from point A to point B. So changing can pick. Energy is equal to 7.5 minus 1.2. The magnitude of the world that is equal to see explain they jewelers. This is that required answer

Jayashree Behera · Answer

so there is no fiction on the hill. Therefore, the amount off wall done by the force f pulling the block uphill is equal to the change in potential energy off the block. And this is regardless off the path taken by the block so we can drive the potential energy to be mass times gravity times the height, the changing height off the block. So substituting the values for moss gravity and the change in height of the distance traveled by the block 40 girlie. So that is two meter. So we get the work done by the fourth F pulling the block up here to be equal to 40. Jules, the correct answer is option e. Thank you.

Kristela Garcia · Answer

in this problem, you are shooting a rocket with a mass of five kilograms and it&#x27;s traveling from an initial velocity of 10 meters per second to a final velocity of 15 meters per second. And it tells us that the work done by the force to go from 10 meters per second to 15 meters per second. So the work done by the supplied force is 2000 jewels of work, so we can apply the work energy through, um that tells us that the network done is equal to our change in kinetic energy. Now, when you&#x27;re doing changing kinetic energy, it&#x27;s super important that you do final kinetic energy minus initial. This is our net work. So when we&#x27;re solving for a network, it&#x27;s the work done by all of the forces on our object. So we can calculate this to give us and I can actually factor out a 1/2 em here. So this is 1/2 em, and I&#x27;m left V squared minus B not squared into 1/2 five 15 squared, minus 10 squared and gives me a positive network done of 312.5 jewels. Okay, again, this is net work. So our work done is with the work done by the Applied Force, plus the work done by gravity. Now we know the work done by gravity is a negative value because it is trying to slow my object down. And we can say that this work done by the Applied Force was 2000. So then the work done by gravity is 312.5 minus 2000 to get me negative. 1687.5 Jules.

Problem

A student pushes a 6 kg box up an inclined plane …

Problem 22 Easy Difficulty

Answer

Related Courses

Cracking The AP Physics 1 Exam

Related Topics

Discussion

Top Educators

Recommended Videos

Watch More Solved Questions in Chapter 14

Video Transcript

Robert Franek

Cracking The AP Physics 1 Exam

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Cracking The AP Physics 1 Exam

Robert FranekCracking The AP Physics 1 Exam

Robert Franek

Cracking The AP Physics 1 Exam