|___45m_____| |___10m____| Kelly looks for an old mine cart that would take the

Question

|___45m_____|                                                                |___10m____|

Kelly looks for an old mine cart that would take the samples and herself to the back entrance of the mine. Knowing the layout of the mine track, she knows she needs to push the cart 45 m, jump in, and have enough speed to get over the 6 m hgih hill (assume there are no frictional forces slowing her down).

1) the greatest force she can push is 890 N. Assuming the combine mass of Kelly and the mine cart if 3800 kg, what speed can she give the cart after pushing it for 45 m? Assume she is pushing horizontally.

2) Using the speed you just calculated, how fast will she be going over the hill?

3) Kelly is now at the bottom of the hill and she notices the track ends at a cave wall near the exit. She applies the breaks and the wheels lock, skidding and sparking all the way down the track. She has 10 meters to stop or she will crash into the wall. The coefficient of kinetic friction between the wheels and track is 0.33. Will she be able to stop in time or will she crash into the wall? Prove answer by finding out how far she will travel?

Explanation / Answer

a) Use F = ma to find the acceleration of the cart which is .234 m/s^2. Then the kinematic equaition d= 1/2 * a * t^2 + Vi*t^2, where a = .234 and Vi= 0 and d = 45m. Solve for t. t = 19.6 s. Then a*t = Vf in this case, so Vf = 4.58 m/s. b)Use PE + KE =MEtotal to find the energy right before the first drop. PE = mgh, KE = .5mv^2 so MEtotal for the system will be 226055 J or 2.26 x10^5 J. Since Energy is always conserved here, at the top of the 6m hill use the same mgh + 1/2mv^2 = 2.26x10^5 and solve for v. You get v = 1.17 m/s c) Do the same thing as in part b to determine her speed. So you get v = 7.56 m/s. Then find the Force of Friction (Ff) with the equation Ff=-.33*Fn (Fn is normal force and .33 is the coefficient of kin friction) Normal force is 3800*9.8=37240. Ff=-12289 N. Now you can use the Work Energy Theorem (Work = Change in Kinetic Energy). So here you have her KE at 108591 J. To find out the stopping distance use W = Fd ==> 108591 = (12289)(d). d = 8.84 meters, so she will stop in time

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