I won’t make fun of you.
“Calculate the speed of the sled with the child after she jumps onto the sled.”
Momentum is conserved; the total momentum of the system (child + sled) is the same before and after the child jumps on the sled.
Before the child jumps on the sled, the momentum of the child is (p = mv) 50kg * 6m/s = 300kgm/s, and the momentum of the block is 10kg * 0m/s = 0kgm/s. So the total momentum of the system is 300kgm/s + 0kmg/s = 300kgm/s.
This total momentum is conserved. Before and after the child jumps on the sled, the total momentum must equal 300kgm/s.
After the child jumps on the sled, the momentum is p = (child mass + sled mass) * velocity, or p = (50kg + 10kg) * v = 60 * v.
Using momentum conservation, we set the momentum before the event equal to the momentum afterwards. 300kgm/s = 60 * v —> v = 5m/s
“Calculate the kinetic energy of the sled with the child after she jumps onto the sled.”
Kinetic energy obeys KE = (½)mv^2. Assuming they question is asking for the kinetic energy of the total system (child + sled), simply substitute the known values. KE = (½)(60kg)(5m/s)^2 = 750J.
“How much work must be done by friction to bring the sled with the child to a stop?”
From the Work-Kinetic Energy Theorem, we know that the amount of work done on a body is equal to the change in kinetic energy of the body. The system starts with 750J, and ends with 0J (v = 0), so the change in KE, and thus the work done by friction to stop the system, is 750J. The friction force give here seems to be irrelevant, although you could use it to determine how far the system slides before it stops.