Fastest snow sled for snow hill?

A 46.5 kg sled and person are moving along a level section of snow elevated at a height of 5.0 meters. The spe?

• A 46.5 kg sled and person are moving along a level section of snow elevated at a height of 5.0 meters. The speed of the sledder is 13.1 m/s. The sled then reaches a hill inclined at 30-degrees. Upon reaching the bottom of the hill, the sledder slows to a stop across a horizontal section of snow. The coefficient of friction along the incline and the horizontal section in the diagram above is 0.21. Determine the distance the sledder slides along the horizontal section before coming to a stop.

• Answer:

  ENERGY AT THE TOP = ENERGY AT THE BOTTOM, FIND THE SPEED AT THE BOTTOM 1/2*m*v1^2 + m*g*h1 = 1/2*m*v2^2 + m*g*h2 In this case, set h2 = 0. All the m's cancel 1/2*v1^2 + g*h1 = 1/2*v2^2 v2 = sqrt(v1^2 + 2*g*h) Find the deacceleration Sum of the forces in the direction of motion = m*a = -friction Sum of the forces in the vertical direction = 0 = N - m*g N = m*g friction = u*N = u*m*g m*a = -u*m*g a = -u*g Now you know the initial velocity (v2 from the previous answer), final velocity (0 m/s), and acceleration vf^2 = vi^2 + 2*a*d d = (vf^2 - vi^2) / (2*a) = (0 - (v1^2 + 2*g*h)) / (-2*u*g) d = (v1^2 + 2*g*h) / (2*u*g)