Please only answer the exercise question at the bottom EXAMPLE 5.5Platform DiverGOAL Use conservation of energy to calculate the speed of a bodyfalling straight down in the presence of gravity.10.0 mKE; = 0PE,- mgy;PROBLEM A diver of mass m drops from a board 10.0 m above thewater's surface, as in the figure. Neglect air resistance. (a) Useconservation of mechanical energy to find his speed 5.00 m above thewater's surface (b) Find his speed as he hits the water.5.00 m-STRATEGY Refer to the problem-solving strategy. Step 1: Thesystem consists of the diver and Earth. As the diver falls, only theforce of gravity acts on him (neglecting air drag), so the mechanicalKE, = mu,?energy of the system is conserved, and we can use conservation ofenergy for both parts (a) and (b). Step 2: Choose y = 0 for thePE, =0water's surface. Step 3: In part (a), y = 10.0 m and y = 5.00 m arethe points of interest, while in part (b), y = 10.0 m and y = 0 m are of interest.SOLUTION(A) Find the diver's speed halfway down, at y = 5.00 m.Step 4: We write the energyKE, + PE, = KE, + PE,conservation equation and supply the+ mv? + mgy;+ mgy =proper terms.Step 5: Substitute v; = 0, cancel the0 + gy, -v? + gy,mass m, and solve for v,.v, = V 2g(y, - v) = V 2(9.80 m/s?)(10.0 m - 5.00 m)V = 9.90 m/s(B) Find the diver's speed at the water's surface, y = 0.Use the same procedure as in part (a),taking y- 0.0 + mgy, =mv? + 0Vr -V 2gy, - / 2(9.80 m/s²)(10.0 m) = 14.0 m/sLEARN MOREREMARKS Notice that the speed halfway down is not half the final speed. Another interesting point isthat the final answer doesn't depend on the mass. That is really a consequence of neglecting the changein kinetic energy of Earth, which is valid when the mass of the object, the diver in this case, is muchsmaller than the mass of Earth. In reality, Earth also falls towards the diver, reducing the final speed, butthe reduction is so minuscule it could never be measured.QUESTION Qualitatively, how will the answers change if the diver takes a running dive off the end ofthe board?O Answer (a) would increase and answer (b) would decrease.O Answer (b) would increase and answer (a) would decrease.Both answers would decrease.O Both answers would increase.O Both answers would remain the same.PRACTICE ITUse the worked example above to help you solve this problem. A diver of mass m drops from a board 8.00m above the water's surface, as shown in the figure. Neglect air resistance.(a) Use conservation of mechanical energy to find his speed 4.00 m above the water's surface.8.854m/s(b) Find his speed as he hits the water.v m/s12.522EXERCISEHINTS: GETTING STARTED I "M STUCK!Use the values from PRACTICE IT to help you work this exercise. Suppose the diver vaults off thespringboard, leaving it with an initial speed of 3.04 m/s upward. Use energy conservation to find his speedwhen he strikes the water.m/s

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Description: Please only answer the exercise question at the bottom EXAMPLE 5.5Platform DiverGOAL Use conservation of energy to calculate the speed of a bodyfalling straight down in the presence of gravity.10.0 mKE; = 0PE,- mgy;PROBLEM A diver of mass m drops fro

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Step 5: Substitute v, = 0, cancel the 0 + gy, -v? + gy, mass m, and solve for v,. v, = V 2g(y, – Y) = V 2(9.80 m/s?)(10.0 m – 5.00 m) Vr = 9.90 m/s (B) Find the diver's speed at the water's surface, y = 0. Use the same procedure as in part (a), 0 + mgy, =mv? + o v - V 2gy, - V 2(9.80 m/s?)(10.0 m) = 14.0 m/s taking y,= 0. LEARN MORE

Step 5: Substitute v, = 0, cancel the 0 + gy, -v? + gy, mass m, and solve for v,. v, = V 2g(y, – Y) = V 2(9.80 m/s?)(10.0 m – 5.00 m) Vr = 9.90 m/s (B) Find the diver's speed at the water's surface, y = 0. Use the same procedure as in part (a), 0 + mgy, =mv? + o v - V 2gy, - V 2(9.80 m/s?)(10.0 m) = 14.0 m/s taking y,= 0. LEARN MORE

College Physics

10th Edition

ISBN:9781285737027

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter5: Energy

Section: Chapter Questions

Problem 41P: (a) A child slides down a water slide at an amusement park from an initial height h. The slide can...

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Alex and John are loading identical cabinets onto a truck. Alex lifts his cabinet straight up from the ground to the bed of the truck, whereas John slides his cabinet up a rough ramp to the truck. Which statement is correct about the work done on the cabinet-Earth system? (a) Alex and John do the same amount of work, (b) Alex does more work than John, (c) John does more work than Alex, (d) None of those statements is necessarily true because the force of friction is unknown, (e) None of those statements is necessarily true because the angle of the incline is unknown.
Answer yes or no to each of the following questions. (a) Can an objectEarth system have kinetic energy and not gravitational potential energy? (b) Can it have gravitational potential energy and not kinetic energy? (c) Can it have both types of energy at the same moment? (d) Can it have neither?
How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and kilocalories.
Review. You can think of the workkinetic energy theorem as a second theory of motion, parallel to Newtons laws in describing how outside influences affect the motion of an object. In this problem, solve parts (a), (b), and (c) separately from parts (d) and (e) so you can compare the predictions of the two theories. A 15.0-g bullet is accelerated from rest to a speed of 780 m/s in a rifle barrel of length 72.0 cm. (a) Find the kinetic energy of the bullet as it leaves the barrel. (b) Use the workkinetic energy theorem to find the net work that is done on the bullet. (c) Use your result to part (b) to find the magnitude of the average net force that acted on the bullet while it was in the barrel. (d) Now model the bullet as a particle under constant acceleration. Find the constant acceleration of a bullet that starts from rest and gains a speed of 780 m/s over a distance of 72.0 cm. (e) Modeling the bullet as a particle under a net force, find the net force that acted on it during its acceleration. (f) What conclusion can you draw from comparing your results of parts (c) and (e)?
“ E=K+Uconstant is a special case of the work energy theorem.” Discuss this statement.