How high a hill can a car coast up (engine disengaged with negligible friction) if it has an initial speed of?
102121
Drew, this is another conservation of energy calculation.  Just like the other one I responded to.  So I'm sure you can get it now but I'll go over this one with you.  so the inital speed after dimensional anaylsis is 27.78 m/s (remember there is 3600s in an hr and 1000m in a km).  So using 1/2mv^2 = mgh (by the way the masses cancel).  you will get a total height of 39 meters (2 sig figs!).. Hope this helps
841789
Assume s=275, v=43, r=0.21 & t is time from start to stop; then v=a*t & s=a*t^2/2, ‘a’ being deceleration. Therefore t=v/a, s=a*(v/a)^2/2; thus 2*s*a=v^2 or a=v^2/(2*s).\nAngular deceleration w’=a/r.\nNow I=0.175, so frictional torque T=I*w’. plug numbers in.
133048
Hello, \n\nThis problem is a bit more involved than you might initially think. There are 4 major considerations you must take into account. First, the acceleration is dependant on the angle of the hill at any given time. Therefore, acceleration is NOT constant, rather fluctuating with the incline. Second, the wind resistance is NOT constant by any means. It increases exponentially with the speed of the sled. Third, the force of the kinetic friction is not constant either. While the coefficient of friction does increase somewhat with speed (you could use an average for this), you cannot neglect the increase in friction when the sled goes around a corner. As you know, the friction is based on the normal force of the sled. At 90mph, the centripetal acceleration around a curve is V^2/r (V=velocity, r is radius of corner) which becomes quite considerable, even though steel on ice only has a kinetic coefficient between .01 - .05. \n\nNow that I have bored you out of your mind, the full equation would look something like this (I’m writing everything longhand so you know what I’m talking about)\nSum of the forces = mass x gravity x sin(angle of hill) – [(coefficient of friction) x mass x gravity x cos(angle of hill) – mass x velocity ^2 / radius of curve] – (Force of air drag)\n\nNotice the term included to account for curves: – mass x velocity ^2 / radius of curve\nDrops to zero when they are going in a straight line (r = infinity). \n\nTo find your acceleration (at a specific point in time), you simply divide everything by the mass of the sled and there you go!\n\nPS. don't take physics advice from dental students ;P
216917
you are shifting into a lower gear, it goes  1,2,3,then overdrive to save on gas but has less power, so when you go up a hill it will shift into 3 on its own for more power , when you do that you will use more gas, it makes your motor work a little harder but all in all it causes a little more wear but over a long time ,, just let the car do it on its own.
125809
The struggle here is that you provide too little information to recognize this problem as coherent.\n\nWe may assume that the acceleration of gravity is 9.8 m/s/s or 32 ft/sec/sec.\n\nThe velocity(V) of an object is equal to its initial velocity(v) plus its acceleration(a) times time(t). \nV = v + at or V = v - 32t (in feet per second)\n\nYour object will reach maximum height when the velocity is zero or when the initial velocity is equal to 32 ft/sec/sec * time. With your list of times, you'd just multiply by 32 to solve for the initial velocity, if that is what the problem is actually looking for. There are many other permutations of this data, but hopefully that is a start for you.\n\nGood luck.
789497
Between 30 and 40 mph.
861510
The way to solve this problem is to decompose the forces acting on the car, but not in the x and y direction. Instead, decompose the forces in the direction of the incline and the normal force.\n\nHence, we have:\n\nF(friction) = μN, moving up the incline, and μ is the coefficient of static friction\n\nThen we have gravity acting on the car, and the component that's acting down the incline is = mgsinθ, where θ is the incline of 15°.\n\nFor the car to be safe in making this turn, the forces along the incline must be equal:\n\n(a) F(friction) = μN = mgsinθ = (1450)(9.8)sin(15°) = 3677.82 N\n\nBut the forces acting perpendicular to the incline on the car must also be zero:\n\nN = mgcosθ\n\nNow substituting this into the previous equation, we get:\n\nμN = μ(mgcosθ) = mgsinθ, this implies that the coefficient of static friction is:\n\n(b) μ = sinθ/cosθ = tanθ = tan(15°) = 0.268
923574
It goes into thermal energy (heat) as the brakes heat up due to friction and perhaps to other, less significant sources of energy loss (e.g. sound of brakes squealing).\n\nNote: The car doesn't really gain energy, it just converts its gravitational potential energy (from being at the top of the hill) into kinetic energy (the energy associated with motion).
59101
It actually would be possible, yes.  The momntum of the car doesn't really change when it drives onto the ramp.  if the ramp is smooth enough then the tires will just spin and the car won't launch through the roof.  If the ramp has enough traction then this trick would be hard on the transmission would probabaly kill a stick shift, but again the car would not launch through the roof.
981896
The upward reaction force (at the top of the bump) is NOT mg + mv^2/r.\n\nPer unit mass, it is g - v^2/r. Note the sign.\n\nAlso, the relevant mass is not the whole mass of the car m but, say, about *half* of it---the rear wheels are not even in contact with the bump unless it's a really huge one (was this overlooked when the question was set?)\n\nThe previous answers are *almost* correct---except for the sign mistake (+ instead of - in the centripetal term). Taking the upward direction as +ve, the gravity force is -ve (-g per unit mass), the bump's reaction n is +ve, and the centripetal force is -ve (c = -v^2/r per unit mass), with c = -g + n (gravity and reaction add up to provide the req'd centripetal acceleration, which is negative, i.e. downward). So, per unit mass, the reaction is \n\nn = -gravity + centripetal = g - v^2/r.\n\nIOW, at that point, the bump's curvature actually lessens the load on the wheels (it's a different story at the point of initial contact of course). \n\nThe relevant mass for the front wheels is m/2, so the numerical answer is:\n\nReaction on the *front* wheels is \n(g - v^2/r)m/2 = (9.8 - (16.6)^2/42.1)1720/2\n= 2800 (newtons) = 286 kG (kilograms-force)\n\n[An answer that forgets to halve the mass would be 5600 N.]\n\nReaction on the *rear* wheels is: gm/2 = 9.8*1720/2 = 8428 (newtons) = 860 kG (kilograms-force)\n\n\nPS: I see now that the correct sign did also occur to gp4rts in his last edit.
834291
Since the first incline is frictionless, there is no energy loss in coming down. Hence KE at the foot of the incline = PE at the start. This gives the velocity at the foot of the first incline as\n1/2 m*v^2 = m*g*h.\nThe frictional force on the horizontal surface = Uk*m*g\nThe work done in overcoming the fricional force =Uk*m*g*L\nThus the available energy at the foot of the second incline = m*g*h - Uk*m*g*L = m*g*(h - Uk*L)\nIf the height to which the block rises is h1,\nm*g*h1 = m*g*(h - Uk*L). Hence \nh1 = h - Uk*L
368139
If the object has onboard power, then the speed is not really relevant to whether it will leave the atmosphere, just when. If you just fire an object upwards and it has no self-propulsion, however, then you need to calculate the change in gravitational potential energy (find out how far the atmosphere extends) and add the energy lost to air friction (this depends on velocity). Equate this to the object's initial kinetic energy and solve for initial velocity v. Your initial velocity will have to be larger than this.
841492
a)The acceleration a = Force / mass = -6070 / 2600 = -2.33 m/s^2\nInitial velocity u = 34.9 m/s\nThe velocity at any time is given by v = u+a*t = 34.9 - (2.33*1.76) = 30.8 m/s\nb) v^2 - u^2 = 2*a*s, where s is the distance travelled in that time. \ns = ( v^2 -u^2)/2a = 57.8 m\nc) The car stops when u + a*t = 0.\nt = 34.9/2.33 = 15 s
782291
Assuming that the friction with the ground provides a constant force, the clumped-together vehicles decelerate at constant rate until they stop. Because we assume the ground to be horizontal, that rate is (mu) * g, where (mu) is the coefficient of friction. Therefore the rate of deceleration is .923 g or 9.052 m/s^2.\n\nThe distance traveled is 18 m. For constant acceleration (or deceleration, it does not matter as long as the velocity at one end of the time period is zero) x = 1/2 a t^2. We know x and a; therefore we find t = 1.994 sec.\n\nNow we use this time and the assumption of constant acceleration to find that the initial velocity was v = a * t = 18.05 m/s. Therefore the intial momentum was 18.06 * 3000 = 54 150 kg m/s, directed toward the northwest.\n\nTo cause that momentum, the momenta of the individual vehicles added, and they are vectors. The truck was headed north, the car, west, and they ended up going northwest. Therefore they had equal-magnitude momenta. (The tangent of the resultant angle is the ratio of the two.)\n\nSay Pt = truck momentum = Vt Mt, Pc = Vc Mc.\nDealing just in magnitudes for the moment, \nPt = Pc --> Vt Mt = Vc Mc, so Vc = Vt (Mt / Mc) = 2.75 Vc.\nBecause the vehicles started at right angles to each other, the magnitude of the final momentum is Pf = sqrt(Pt^2 + Pc^2) = \nsqrt((Mt * Vt)^2 + (Mc * 2.75 * Vt)^2) = 3111 * Vt\nWe have already found Vf = 18.05, and we know the combined mass is 3000 kg; therefore we infer Vt = 18.05 * 3000 / 3111 = 17.41 m/s. Therefore Vc = 47.87 m/s.
698763
Vy=sin(36)(30)=17.63m/s\nVx=cos(36)30= 24.27m/s\n\nwhen it reaches the ground, the speed will be the same as Vy\nVf=at+Vi\n-17.63m/s=(-9.8m/s^2)t+17.63\n-35.26m/s=(-9.8m/s^2)t\nt= 3.597 s\n\nwhen it reaches the highest point, its velocity will be 0m/s\nVf^2=2ad+Vi^2\n(0m/s)^2=2(-9.8m/s^2)d+(17.63m/s)2\n-310.81m^2/s^2=(-19.6m/s^2)d\nd= 15.85m\n\nx=vt\nx=(24.27m/s)(3.597s)\nx=87.29m
891305
In the absence of friction, the acceleration of an object down a 12-degree slide would be\n\n9.81 * sin 12, or around 2.04\n\nInstead, your obejct is accelerating at 1.22.\n\nThe loss of acceleration due to friction is\n\n(9.81 * sin 12) - 1.22\n\nand the coefficient of friction is thus\n\n[(9.81 * sin 12) - 1.22] / (9.81 * sin 12)
757156
How about someone is run over by a car.  The owner of the car says they left it in neutral on a level parking lot, and must have forgot to set the parking brake -- the car started rolling forward before they could jump back in and stop it, and the car ran the victim over.\n\nThis can be disproven because Newton's first law (the "Law of Inertia") states that an object at rest will stay at rest until acted on by an outside force.  Since no outside force was present (the parking lot was level so no gravitational force acts to accellerate the car horizontally) and the car itself was turned off, it cannot possibly have "just started to roll by itself" -- the car owner intentionally ran the victim over.\n\nWhy would he do that?  Well I guess that will be the fun and imaginitive part of the assignment!