A boy and a sled of combined mass 80 kg start at rest on the top of n hill 15 m

Question

A boy and a sled of combined mass 80 kg start at rest on the top of n hill 15 m on a winter day at 0 degrees Celsius. He sleds down the hill with no friction. What is the speed of the boy and his sled at tire bottom of the hill? The boy notices he is heading for a cliff 20 m ahead and applies the breaks, producing a coefficient of kinetic friction between the sled and the ice of mu_k = 0.8. How far does the boy slide before stopping, or does he fall over the cliff? As he slows, all his kinetic energy is converted by friction into heat, which serves to melt the ice he is sliding over. What mass of ice is melted during this process?

Explanation / Answer

Combined mass of the boy and sled m = 80 kg

Height of the hill h = 15 m

a)Let the velocity of the system at the bottom be v

The potential energy of the system converts to kinetic energy by the time it reach the bottom

So potential energy U = kinetic energy K

   We can write m * g * h = ½ * m * v2

                      80 * 9.8 * 15 = ½ * 80 * v2

                                        v2 = ( 9.8 * 15 * 2 )

                                        v2 = 294

                                        v   = 17.14 m/s

so the velocity of the sled at the bottom is v = 17.14 m/s

b) Let the stopping distance be S

coefficient of kinetic friction µk = 0.8

stopping distance is given by S = v2/( 2 * µk * g )

                                                         = 294/ ( 2 * 0.8 * 9.8 )

                                                         = 18.75 m

So the boy and sled does not fall over the cliff

c) since the boy and sled system comes to rest finally , we can take the change in the kinetic energy equal to initial kinetic energy

this must be equal to heat energy supplied to ice

kinetic energy of the boy sled system K = ½ m * v2

                                                                         = ½ * 80 * 294

                                                                         = 11760 J

Let the mass of the ice be mice

Latent heat of fusion L = 335000 J/kg

Let Q be the energy supplied to ice , then

                          K = Q

                        11760 = mice * L

                         11760 = mice * 335000

                             mice = 11760 /335000

                                     = 0.035 kg

So the mass of the ice converted is very less which is equal to mice = 0.035 kg

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