# Problem Set 5 Work and Kinetic Energy Solutions

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department o Physics Physics 8.1 Fall 1 Problem Set 5 Work and Kinetic Energy Solutions Problem 1: Work Done by Forces a) Two people push in opposite directions on a block that sits atop a rictionless surace (The soles o their shoes are glued to the rictionless surace). I the block, originally at rest at point P, moves to the right without rotating and ends up at rest at point Q, describe qualitatively how much work is done on the block by person 1 relative to that done by person? Solution: Initially the block is at rest. Ater the pushing has ended, the block ends at rest, so the change in kinetic energy is zero. From the work-kinetic energy theorem, this implies that the total work done on the block is zero. The total work done on the block is the sum o the work done on the block by each person. Since the block moves to the right, person who pushes the block to the right does a positive work, and the person pushing to the let does negative work. Since the total is zero, the magnitude o the work done by each person is equal. b) Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same orce to stop each. How do the distances needed to stop them compare? Explain your reasoning. Answer 3. The kinetic energy o an object can be written as K = p / m. Because the ping pong ball and the bowling ball have the same momentum, the kinetic energy o the less massive ping pong ball is greater than the kinetic energy o the more massive bowling ball. You must do work on an object to change its kinetic energy. I you exert a constant orce, then the work done is the product o the orce with the displacement o the point o application o the orce. Since the work done on an object is equal to the change in kinetic energy, the ping pong ball has a greater change in kinetic energy in order to bring it to a stop, so the you need a longer distance to stop the ping pong ball.

2 Alternative Solution: Both the initial momentum and the orce acting on the two objects are equal. Thereore the initial velocity and the acceleration o the ping-pong ball isgreater than the bowling ball by the ratio o the bowling ball mass to the ping-pong ball mass. m v = m v = p p, p b, b x mpap = mbab = Fx Since both the orce acting on each object and the change in momentum is the same, the impulse acting on each ball is the same. Thereore, the time interval it takes to stop each object is the same. Since the displacement is equal to " ax! t #! x = % vx, \$ &! t ' ( The ratio! xp m = b! x m hence the ping-pong has the greater displacement. b p

3 Problem : Kinetic Energy and Work A tetherball o mass m is attached to a post o radius b by a string. Initially it is a distance r rom the center o the post and it is moving tangentially with a speed v. Ignore gravity and any dissipative orces. a) Suppose the string passes through a hole in the center o the post at the top and is gradually shortened by drawing it through the hole (igure above let). Until the ball hits the post, is the kinetic energy o the ball constant? Explain your reasoning. Answer: Since the path o the ball is not circular, a small displacement o the ball has a radial component inward so the dot product between the orce by the rope on the ball with the displacement is non-zero hence the work done by the orce is not zero. Thereore the kinetic energy o the ball changes.. b) Now suppose that the string wraps around the outside o the post (igure above right). Until the ball hits the post, is the kinetic energy o the ball constant? Explain your reasoning.

4 Answer A small displacement o the ball is always perpendicular to string since at each instant in time the ball undergoes an instantaneous circular motion about the string contact point with pole. Thereore the dot product between the orce by the rope on the ball with the displacement is zero hence the work done by the orce is zero. Thereore the kinetic energy o the ball does not change.

5 Problem 3: An object o mass m = 4. kg, starting rom rest, slides down an inclined plane o length l = 3.m. The plane is inclined by an angle o! = 3 to the ground. The coeicient o kinetic riction µ k =.. At the bottom o the plane, the mass slides along a rough surace with a coeicient o kinetic riction µ k (x) = (.5 m!1 )x until it comes to rest. The goal o this problem is to ind out how ar the object slides along the rough surace. a) What is the work done by the riction orce while the mass is sliding down the inclined plane? Is this positive or negative? b) What is the work done by the gravitational orce while the mass is sliding down the inclined plane? Is this positive or negative? c) What is the kinetic energy o the mass just at the bottom o the inclined plane? d) What is the work done by the riction orce while the mass is sliding along the ground? Is this positive or negative? e) How ar does the object slide along the rough surace? Solution: a) While the object is sliding down the inclined plane the kinetic energy is increasing due to the positive work done on the object by the gravitational orce and the negative work (smaller in magnitude) done by riction orce. As the object slides along the level surace, the (negative) work done by the riction orce slows the object down. We will use the work-kinetic energy theorem to calculate the change in kinetic energy or each stage. The ree body diagram on the inclined plane and on the level surace are shown below.

6 Choose a coordinate system with the origin at the top o the inclined plane and the positive x-direction pointing down the inclined plane. Then the work done by the riction orce is x = l!! W = F " dr = F dx = #( µ ) Nl = #( µ ) mg cos! l < riction \$ \$ x = x k inclined k inclined W riction =!(.)(4.kg)(9.8m "s - )(3.m)(cos(3 o ) =!.4 J Note that the normal orce is determined rom Newton s Second Law applied to the normal direction to the inclined plane. N! mg cos" = b) The work done by the gravitational orce is just W grav =!mg(h! h ) note that (h! h ) = -l sin". So the work done by the gravitation orce is The magnitude o this work is W grav = mgl sin! > c) The total work is W grav = (4.kg)(9.8 m! s - )(3.m)(sin(3 o ) = 58.8 J W total = W grav + W riction = mgl(sin! " (µ k ) inclined cos!) W total = 58.8 J ".4 J = 38.4 J The change in kinetic energy is just equal to the inal kinetic energy at the bottom o the incline because the started rom rest,!k = 1 mv bottom So the work-kinetic energy theorem W total =!K becomes mgl(sin! " (µ k ) inclined cos!) = 1 mv bottom = 38.4 J e) Choose a coordinate system with the origin at the base o the inclined plane and the positive x-direction pointing in the direction the object moves along the plane. The

7 normal orce on the object is determined rom Newton s Second Law applied to the normal direction to the plane. N! mg = Then the work done by the riction orce is W riction = " F!! d r! x = d x = d = " F x dx = " (.5 m #1 )Nx dx x = W riction = #(.5 m #1 )N d x = = #(.5 m#1 )mg d < ) The change in kinetic energy is just equal to the kinetic energy at the bottom o the incline plane because the object comes to rest!k = " 1 mv bottom So the work-kinetic energy theorem W total =!K becomes! 1 mv =!(.5 bottom m!1 )mg d However we have already determined this kinetic energy so this last equation becomes!mgl(sin"! (µ k ) inclined cos") =!(.5 m!1 )mg d we can now solve or the distance it traveled beore it came to rest along the horizontal d = (mgl(sin! " (µ k ) inclined cos!)) (.5 m "1 )mg = (38.4 J) (.5 m "1 )(4.kg)(9.8m # s - ) = 6.3 m Note: I we consider the entire motion rom the release at the top o the inclined plane to coming to rest on the horizontal, then the total change in kinetic energy is zero and Thus =!K = W total = W grav + (W riction ) inclined + (W riction ) horizontal = mgl sin! " mgl(µ k ) inclined cos! " (.5 m "1 )mg d which leads to the identical expression or the distance traveled on the horizontal beore coming to rest.

8 Problem 4: Asteroid about Sun An asteroid o mass m is in a non-circular closed orbit about the sun. Initially it is a distance r i rom the sun, with speed v i. What is the change in the kinetic energy o the asteroid when it is a distance is r, rom the sun? Solution: We shall use the work-kinetic energy theorem, 1 1 W =! K " K # K = mv # mv i i The work done by the gravitational orce is the line integral Let s choose polar coordinates. W!! =!. r " F r g d r i The gravitational orce between the sun and the asteroid is given by! F Gm =! r sun g m ˆ r. As the asteroid moves, the ininitesimal displacement is tangent to the path and is given in polar coordinates by

9 ! dr = dr ˆr + rd! èˆ The work done by the gravitational orce on the body is given by the line integral r!! r Gmsunm W = g ( ˆ \$ F " dr = # ˆ " dr ˆ + rd! ) r \$ r r è. i ri r Note that because ˆ r! è ˆ = and ˆr! ˆr = 1, only the radial part o the displacement contributes to the work done by the gravitational orce, W r ri Gm r sun m dr =!". Upon evaluation o this integral, we have or the work W = # dr = = Gm! m # " & ' r ( r Gmsunm Gmsunm 1 1 r sun i r r \$ r r i r %. i Using the work-kinetic energy theorem, the change in kinetic energy is! 1 1 " # K = Gmsunm \$. % r r & ' i ( The inal kinetic energy o the asteroid is then 1 1! 1 1 " mv = + mvi + Gmsunm # \$ r r % & i ' Let s check our result by considering the case that the asteroid is moving closer to the sun, r < ri, hence 1/ r > 1/ ri. Thus the work done by gravitational orce on the asteroid is positive,! 1 1 " W = Gmsunm # >. \$ r r % & ' and the kinetic energy o the asteroid increases as we expect or a body moving closer to the sun.

10 Problem 5 Work Done by a Several Forces A block o mass m slides along a horizontal table with speed v. At x = it hits a spring with spring constant k and begins to experience a riction orce. The coeicient o riction is given by µ. How ar did the spring compress when the block irst momentarily comes to rest. Solution: From the model given or the rictional orce, we could ind the non-conservative work done, which is the same as the loss o mechanical energy, i we knew the position x where the block irst comes to rest. The most direct (and easiest) way to ind x is to use the work-energy theorem, x= x \$ x= 1 1 F dx =! K " K # K = mv # mv x i Since we are trying to ind the distance that the object moved when it irst becomes to rest, we have that v =, so the work-kinetic energy theorem becomes x= x 1! Fx dx = mv. x= There are two orces acting on the block, riction and the spring orce. The x-component o these orces are given by F = F + F =! kx! µ mg. x x,spring x, riciton So the work done on the object is x= x x= x x= x " " " W = F dx = F dx + F dx x x,spring x,riction x= x= x= x x =!" kx dx! " µ mg dx (1) 1 =! kx! µ mgx

11 Applying the work energy theorem yields We can rearrange this as 1 1! k x! µ mgx =! mv () µ mg m x + x! v =. k k The solution o this quadratic equation is given by x =! µ mg k ± " # \$ µ mg k % & ' + m k v Note that we have assumed that x >, thereore we need to choose the positive square root, x =! µ mg k + " # \$ µ mg k % & ' + m k v. It is worth checking that the above result is dimensionally correct. Recall that Hooke s law states that F =! kx, so µ mg / k has the dimensions o length. Similarly mv has the dimensions o energy or orce times distance. So m k v has the dimensions o length squared.

12 Problem 6 Sticky Pendulum A simple pendulum consists o a bob o mass m 1 that is suspended rom a pivot by a string o length l and negligible mass. The bob is pulled out and released rom a height h as measured rom the bob s lowest point directly under the pivot point and then swings downward in a circular orbit as shown in the igure below. At the bottom o the swing, the bob collides with a block o mass m that is initially at rest on a rictionless table. Assume that there is no riction at the pivot point. a) What is the work done by the gravitational orce on the bob rom the instant when the bob is released to the instant just beore the collision? b) How much work does the tension orce do as the bob moves in a circular path? c) What is the speed o the bob at the bottom o the swing immediately beore the collision? d) Suppose the bob and block stick together ater the collision. What is the speed o the combined system immediately ater the collision? e) What is the tension in the string immediately ater the collision? ) What is the change in kinetic energy o the block and bob due to the collision? What is the ratio o the change in kinetic energy to the kinetic energy beore the collision? Solution: a) Choose a coordinate system with the origin at the bottom o the swing as shown in the igure below. The gravitational orce is m g! =!mgĵ. The displacement is d r! = dx î + dy ĵ. Thereore the work done by the gravitational orce is W g = " m g!! d r! = " #mgĵ!(dx î + d y\$ ĵ) = #mg " d y\$ = #mg( y # y i ) = mgh i i y \$ = y y \$ = y i

13 b) The tension orce does zero work because it is perpendicular to the displacement o the bob. c) We can use the work energy theorem, W g =!K, noting that the bob started rom rest m 1 g h = 1 m v 1 1, b (3) Thereore the speed o the bob at the low point o the swing just beore the collision, v 1, b = g h. (4) d) Consider the bob and the block as the system. Although tension in the string and the gravitation orce are now acting as external orces, both are particular to the motion o the bob and block during the collision. I we additionally assume that the collision is nearly instantaneous, then the momentum is constant in the direction o the bob s motion, m 1 v 1, b = (m 1 + m )v a, (5) where v a is the speed o the bob and block immediately ater the collision. Thereore Using Eq. (4) in Eq. (6) yields v a = m 1 m 1 + m v 1, b. (6) v a = m 1 m 1 + m gh. (7) d) We show the ree body orce diagram immediately ater the collision. Newton s Second Law becomes v!t + 4m 1 g =!4m a 1 l Thereore substitute Eq. (7) into Eq. (8) and solve or the tension in the string immediately ater the collision (8)! T = 4m 1 g 1+ h # " l m 1 \$ (m 1 + m ) & % (9)

14 e) The change in kinetic energy o the bob and block due to the collision in part c) is given by!k = K ater " K beore = 1 4m v " 1 1 a m v. (1) 1 1,b Using Eq. (7), and (4) in Eq. (1) yields!k = K ater " K beore = 1 4m v " 1 1 a m v 1 1,b!K = m m gh 1 1 (m 1 + m ) " 1 m gh = m gh 1 1 # 4m 1 (m 1 + m ) " 1 & % \$ ( ' (11) The kinetic energy beore the collision was m1gh, and so the ratio o the change in kinetic energy to the kinetic energy beore the collision is!k K beore = # 4m 1 (m 1 + m ) " 1 & % \$ ( '. (1)

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