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Topic: **Work and energy problem solving****Question:**
a 15kg cart is moving with a velocity of 7.50m/s down a level hallway. a constant force of -10N acts on the cart, and its velocity becomes 3.20m/s.
a. what is the change in kinetic energy of the cart?
b. how much work was done on the cart?
c. how far did the cart move while the force acted?

July 19, 2019 / By Bell

A 17.2 kg block is dragged over a coarse, horizontal floor by making use of a relentless tension of seventy 9 N appearing at an attitude of attitude 28.9? above the horizontal. The block is displaced 6.5 m, and the coefficient of kinetic friction is 0.174. discover the artwork achieved by making use of the seventy 9 N tension. The acceleration of gravity is 9.8 m/s2 . The horizontal factor of the seventy 9 N appearing at an attitude of attitude 28.9? above the horizontal = seventy 9 * cos 28.9° = sixty 9.sixteen N The vertical factor of the seventy 9 N appearing at an attitude of attitude 28.9? above the horizontal = seventy 9 * sin 28.9° N lifting up the burden of the 17.2 kg block is the vertical tension pushing down Weight = 17.2 * 9.8 N pushing down the internet vertical tension = 17.2 * 9.8 – seventy 9 * sin 28.9° The friction tension = coefficient of kinetic friction * internet vertical tension The friction tension = 0.174 * (17.2 * 9.8 – seventy 9 * sin 28.9°) The friction tension = 0.174 * (168.fifty six – 38.18) = 22.sixty 9 N as a results of fact the horizontal factor of the seventy 9 N = sixty 9.sixteen N, and the friction tension = 22.sixty 9 N; the internet tension = sixty 9.sixteen – 22.sixty 9 = 40 six.40 seven N The artwork achieved by making use of the internet tension = 40 six.40 seven N * 6.5 M = 302.0.5 N-m The artwork achieved by making use of the seventy 9 N appearing at an attitude of attitude 28.9? above the horizontal = seventy 9 * cos 28.9° * 6.5 = 449.55 N-m artwork achieved by making use of friction = -22.sixty 9 N * 6.5 m = -147.485 N-m discover the artwork achieved by making use of the seventy 9 N tension = 449.55 N-m discover the fee of the artwork achieved by making use of the stress of friction = -147.485N-m discover the artwork achieved by making use of the conventional tension. the conventional tension is perpendicular to the path the block is shifting, so the conventional tension DOES NO artwork!! Even the equation for artwork is of an identical opinion with me artwork = tension * distance * cos ninety° = 0, as a results of fact cos ninety° = 0 what's the internet artwork achieved on the block? 302.0.5 N-m

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Force acting on the box = 120 - mg(sin Θ) - µ(mg)(cos Θ) where m = mass (as defined above) g = acceleration due to gravity Θ = angle of incline = 30 deg Substituting appropriate values (noting that "mg = 61"), F = 120 - 61(sin 30) - 61(0.22)(cos 30) F = 78.94 N Next working formula is Work done to box = Change in potential energy + Kinetic energy Fd = mgh + KE where F = net force acting on the box = 78.94 (as calculated above) d = distance that box was dragged = 8.8 mg = weight of the box = 61 N h = vertical distance that box was dragged = 8.8*sin 30 Substituting appropriate values, 78.94*8.8 = 61(8.8)*sin 30 + KE and solving for KE, KE = 426.272 joules

it somewhat is a conservation of means situation. the entire technique of the skier on the precise of the hill is TE = mgh = PE her means means. h = 40 seven meters, m = seventy 4 kg, and g = 9.80 one m/sec^2 close to Earth's floor. As there's no friction, there's no artwork executed on or by applying the skier as she shusses during the moguls. So, and that's the cool section, her complete means TE keeps to be fastened during the run. yet, and that's the massive yet, the make up of that complete means is consistently changing. right here is why. Her complete means would be written as TE = KE + PE + WE; the place WE = 0 artwork through fact there's no friction. Then on the initiating, on the precise of the hill TE = PE = mgh through fact KE = 0 kinetic means as there's no velocity. remember, her complete means TE = consistent. So on the backside of the hill, the skiers TE = KE + PE + WE = KE through fact PE = mgh = mg0 = 0 while h = 0 and, as in the previous, WE = 0 through fact there's no friction. And we placed the initiating TE = PE = KE = TE on the tip or PE = mgh = a million/2 mv^2 = KE; so as that v^2 = 2gh and KE = PE = mgh. word in v^2 = 2gh we purely derived between the SUVAT equations with the preliminary velocity u = 0. And now you already know, from the physics, the place that got here from to boot.

Let u = velocity at 0 and v = velocity at 2 sec u = (6.0i -2.0j + 4.0k) m/s Let |u| = magnitude of u |u|^2 = 6.0^2 + (-2.0)^2 + 4.0^2 m^2/s^2 = 36 + 4 + 16 m^2/s^2 Or |u|^2 = 56 m^2/s^2--------(1) v = (2.0t^2 i + 4.0t j + 6.0k) m/s for t = 2 Or v = (2.0 * 2^2 i + 4.0 * 2 j + 6.0k) m/s Or v = (8 i + 8 j + 6 k) m/s Let |v| = magnitude of v |v|^2 = 8^2 + 8^2 + 6^2 m^2/s^2 |v|^2 = 164----------(2) Work done W = change in kinetic energy W = 1/2 m|v|^2 - 1/2 m|u|^2 W= 1/2 * (3.0 kg) * (|v|^2 - |u|^2)----------(3) [because mass m = 3.0 kg) Using (1) and (2) in (3) W = 1/2 * 3 * (164 - 56) J = 1/2 * 3 * 108 J = 162 J Ans: 162 J

o.k.. enable's try this! in the starting up, enable's discover the sledge's weight. you likely recognize you may want to multiply the mass by 9.8 by now. So the burden comes out to be 171.5 N. it really is continuously a reliable position to commence :). o.k., now all of us recognize that Fnet will equivalent 0 because it says that it strikes at a consistent % (i.e. acceleration = 0). Fnetx and Fnety will both be 0! the first difficulty we can to not do is discover what the traditional stress is. you may imagine it really is 171.5, even if it really is really no longer! because there's a y area of the stress utilized, the traditional stress will actually be decrease. imagine of it like this - in case you raise something slightly larger from the floor, there'll be a lot less friction. F of friction = M x standard stress. Sounds sturdy, good? and it will make experience too when we placed it contained in the equation: Fnety (keep in mind, strictly contained in the y route now. it really is 0 besides) = FN (standard stress) + Fg (which we realized replaced into -171.5 N earlier!) + FTy (stress of stress, strictly y route) enable's remedy it slightly, shall we? -FN = -171.5 + FTy FN = 171.5 - FTy we do not recognize precisely what the cost of FTy is, yet we do recognize that it really is equivalent to sin(20) * feet (you recognize, the triangle). enable's plug that in: FN = 171.5 - sin(20)*feet Now we may be able to get to the x route. similar difficulty because the Y, enable's get all of our forces: 0 = FTx + Fkf (stress of friction) all of us recognize that Fkf = Mk * FN. We also recognize that Mk = 0.5, so it really is technically 0.5*FN Rearrange back~ 0=FTx + 0.5FN like the y route, we may be able to replace FTx with an expression with feet making use of our triangles. This time, FTx = cos(20) * feet. 0 = cos(20) * feet + (0.5)FN NOW huge PLUGGING IN! keep in mind that we solved for FN earlier: FN = 171.5 - sin(20)*feet Now~ 0 = cos(20) * feet + 80 5.seventy 5 - (0.5)(sin(20))(feet) ANND i imagine I messed something up o_o. answer comes out to be detrimental. likely something incorrect on the maths or something, until eventually... Conspiracy idea - It replaced into actually 2 hundred ranges no longer 20. Sorry about that, yet I had the right line of concept for area a. :) Have a sensible day.

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