Kinetic Energy and Frictional Work of Child Jumping on Sled

[Jfxue]

A 50 kg child running at 6.0 meter per second jumps onto a stationary 10kg sled. The sled is on a level frictionless surface.
1.FIND VELOCITY AFTER CHILD JUMPS ONTO SLED
M1V1+M2V2=(M1+M2)VF
300=60VF
VF=5 M/S

2.FIND KINETIC ENERGY of the sled with the child after she jumps onto the sled.
KE=1/2MV^2
1/2(60)(5^2)=750J

3.After a short time, the moving sled with the child aboard reaches a rough level surface that exerts a constant frictional force of 54 Newtons on the sled. How much work must be done by friction to bring the sled with the child to a stop?

WFf=(Ff)(D) =54d

find d
vf^2=vi^2+2ad
0=50^2 + 2(-54/60)d
1.8d=25
25/1.8 = 13.9M

now i got
wff=54x13.9 =750Jam i correct?

 

[technician]

you have made a mathematical mistake... 300/60 = 5. you have written 50

 

[Jfxue]

ooo but other than that I am correct?

 

[technician]

Yes...looks OK.
The friction part of the question is a bit strange. If you are given the force of friction you are usually asked to calculate the distance (d) to bring the thing to rest.
Are you sure you read the question correctly? because the work needed to bring the sled to rest is always going to be = to the KE of the sled.
A large force of friction means it will come to rest after a short distance (and time)
A small force of friction means a longer distance and time.
Make certain to re-calculate the KE