3207 Shares Topic: Energy problem solving with answer
July 16, 2019 / By Cale
Question: A seasoned mini golfer is trying to make par on a tricky hole #5. The golfer can complete the hole by hitting the ball from the flat section it lays on, up a 45° ramp launching the ball into the hole which is d = 1.00 m away from the end of the ramp. Q: If the opening of the hole and the end of the ramp are at the same height, y = 0.740 m, at what speed must the golfer hit the ball to land the ball in the hole? Assume a frictionless surface (so the ball slides without rotating). The acceleration due to gravity is 9.81 m/s2. The hint I'm given is: "This problem may seem daunting at first. But, the fact that the ball is sliding on a frictionless surface and the ramp being at a 45 degree angle greatly reduce the complexity. Use what you have learned from projectile motion to find an expression for the range in terms of the velocity at the end of the ramp. Then, use the conservation of mechanical energy to solve for the initial speed the ball must have to land into the hole." Tried the method here https://ca.answers.yahoo.com/question/index?qid=20121108221848AAhBYHE And got V=4.132m/s, Which is incorrect. I ws given this hint: "The horizontal range, d, of a projectile off of a 45 degree ramp is given by: d=v^2/g Where, g=9.81m/s^2 and (v) is the velocity of the projectile. Use this equation along with the conservation of mechanical energy to find the speed the ball needs to jump the ramp and land in the hole."  Alfonzo | 10 days ago
Start with the simple hint they give you: d = v² / g v² = g*d = 9.81m/s² * 1.00m = 9.81 m²/s² The KE at the bottom must equal the KE and PE at the top: ½mV² = mgh + ½mv² → mass m cancels ½V² = 9.81m/s² * 0.740m + ½ * 9.81m²/s² = 12.2 m²/s² V = 4.93 m/s ◄ I can't tell what the answerer in your citation has done, except that they've made a mountain of a molehill and done it poorly. Hope this helps!
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We found more questions related to the topic: Energy problem solving with answer Originally Answered: Physics Mechanical Energy Problem?
Paul and Julie are sitting on a sled at the top of hill 1 that is 10 m high vertically with a velocity of 0 m/s. The mass of the sled and its passengers is 100 kg. The sled slides down and up the slope to hill 2 which is 2 m high, before continuing to the bottom of the hill. Calculate the speed of the sled at hill 2 given that the frictional forces acting on the sled between hill 1 and 2 is 3000 J. This is a conservation of energy problem PE = mass * 9.8 * height KE = ½ * mass * velocity^2 Initial KE + PE - Work Initial KE = 0 Initial PE = 100 * 9.8 *10 = 9800 J Work = Work of Friction = 3000 J Total KE + PE – W = 6800 J Final KE + PE Final PE = 100 * 9.8 * 2 = 1960 J Final KE = ½ * 100 * velocity^2 Final energy = Initial energy 1960 + ½ * 100 * v^2 = 6800 50 * v^2 = 4840 v^2 = 96 v = 9.8 m/s Can you tell me if this problem is possible to solve? A pendulum is held next to a ceiling and then released. The pendulum reaches a maximum speed of 3.0 m/s. What is the length of the cord holding the ball? The pendulum’s speed is caused by an increase in its potential energy, which is caused by the change of potential energy. Δ KE = ½ * mass * velocity^2 = ½ * mass * 3^2 = 4.5 * m Δ PE = mass * 9.8 * height Δ KE = Δ PE 4.5 * m = m * 9.8 * Δ height m cancels Δ height = 4.5 ÷ 9.8 = 0.46 m I think you need more information. Time = 2 * π * (L / 9.8)^0.5 Originally Answered: Physics Mechanical Energy Problem?
hi, i'm french. without resistance, the Mechanical ability is an same in the course of the flow. EM = EPotential + EKinetic a million) yet, on the starting up of the flow, the ball has no longer speed so Em = EP. yet Ep = mgz. With m = 0.01, g = 9.80 one and z=2. you may calculate it. 2) even as the ball hits the floor, there is not any ability ability, so Em = Ep + Ek, yet Ep = 0 so Em = Ek. yet, Em is a continuing so Ek = mgz. 3) you may recognize that Ek = a million/2mv² So v=(2Ek/m)^a million/2 I actual have in simple terms defined, to practice the calculations. :) solid success. Tammie
x = v0 [email protected] t fly time is t = 2 v0 [email protected] / g then x = 2 v0^2 [email protected] [email protected] / g and v0 = [(g x) / (2 [email protected] [email protected])]^1/2 v0 = [(9.8*1) / (2*cos45*sin45)]^1/2 = 3.13 m/s
👍 120 | 👎 6 Originally Answered: Work-energy theorem problem?
draw a diagram first of all so you'll comprehend what I've done work = force * displacement in the direction of the force work done on the object = 591* 2.95 / Cos 35 = 1743.45 *0.8192 = 1428.23 J kinetic E = ½mv² initial kinetic energy = 86 * (2.4)² / 2 = 247.68 J gravitation potential is 0 at the bottom of the ramp G. Potential E = mgh final potential energy = 86 * 9.8 * 2.95Sin35 = 2486.26 * 0.5736 = 1426.12 J final: total energy = 1428.23 + 247.68 = 1675.91 J final kinetic energy = total energy - potential energy = 1675.91 - 1426.12 = 249.79 J E = ½mv² 249.79 = 86 * v² /2 v² = 5.81 v = 2.41 m/s that's it Originally Answered: Work-energy theorem problem?
truthfully, gravity IS a drive. Anything that has mass, has gravity. Yes, blackholes suck mild. Light has no mass for the reason that not anything with mass can reap mild velocity. Light does no longer paintings the identical approach as different matters within the universe for the reason that mild and time are warped by means of mass. If there used to be not anything in area, sure, gravity could nonetheless exist. Think of gravity like a blanket stretched tightly via the universe. When a planet is located at the sheet, the sheet bends in a gap across the mass. If there used to be no mass, the "sheet" could nonetheless be there, simply no gap.

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