Aboy shoves his stuffed toy zebra, which has mass m, down a frictionless chute, starting at a height h above the bottom of the chute and with an initial speed of u. the toy animal emerges horizontally from the bottom of the chute and continues sliding along a horizontal surface with coefficient of kinetic friction μ. at what distance d from the bottom of the chute does the toy zebra come to rest? express your answer in terms of the given variables and g, the acceleration due to gravity.

Assumption to make manipulation of units easier. All masses will be in kilogram, distances in meters, and time in seconds. The toy has an initial energy of E = 0.5mu^2 (Units will be kg*m^2/s^2 or Joules) since it starts at a height > 0, it has gravitational potential energy and in the process of falling, that potential energy will be converted to kinetic energy. This energy will be added to the initial energy the toy had. E = mhg (Units will be kg*m^2/s^2 or Joules as well). So the total energy the toy has will be E = 0.5mu^2 + mhg E = m(0.5u^2 + hg) (Units will be kg*m^2/s^2 or Joules) Now this energy needs to be dissipated via kinetic friction. The value for the force that will be slowing down the toy is the normal force multiplied by the coefficient of kinetic friction. And the normal force is the mass of the toy multiplied by the gravitation acceleration. So F = ÎĽmg (Units are kg*m/s^2, or newtons) If you look at the energy the toy has, its units are kg*m^2/s^2 and if you look at the force you have available to slow it down, its units are kg*m/s^2. Those units are almost identical except for an extra unit of m. So if you divide the energy by the force, you'll get a result that's the distance that force has to be applied over. So d = m(0.5u^2 + hg)/(ÎĽmg) (Result will have units of m) Now if you look closely, you'll see that mass is present in both the numerator and denominator. So you can cancel them. Getting: d = (0.5u^2 + hg)/(ÎĽg) (Result will have units of m)