Could you sled backwards down a hill?

You are pulling your sister on a sled to the top of a 18.0 m high, frictionless hill with a 10.0° incline. You?

• You are pulling your sister on a sled to the top of a 18.0 m high, frictionless hill with a 10.0° incline. Your sister and the sled have a total mass of 50.0 kg. You pull the sled, starting from rest, with a constant force of 127 N at an angle of 45.0° to the hill. If you pull from the bottom to the top, what will the speed of the sled be when you reach the top?

• Answer:

  The component of force parallel to the slope (Fc)= 127 x cos45* = 127 x 0.71 = 89.80 Newton The length of the hill(L) = 18/sin10* = 103.66 meter The Net force on the sled Fn = Fc - mgsin10* =>Fn = 89.80 - 50 x 9.8 x 0.17 [ let the acceleration of the sled is a m/s^2] =>m x a = 4.71 =>a = 4.71/50 = 0.094 m/s^2 Let the final velocity of the sled at the top is v m/s, than by :- v^2 = u^2 + 2as =>v^2 = 0 + 2 x 0.094 x 103.66 =>v = sqrt19.54 = 4.42 m/s