Answer:
final velocity will be44.72m/s
Explanation:
HEIGHT=h=100m
vi=0m/s
vf=?
g=10m/s²
by using third equation of motion for bodies under gravity
2gh=(vf)²-(vi)²
evaluating the formula
2(10m/s²)(100m)=vf²-(0m/s)²
2000m²/s²=vf²
√2000m²/s²=√vf²
44.72m/s=vf
Assume the snow is uniform, and horizontal.
Given:
coefficient of kinetic friction = 0.10 = muK
weight of sled = 48 N
weight of rider = 660 N
normal force on of sled with rider = 48+660 N = 708 N = N
Force required to maintain a uniform speed
= coefficient of kinetic friction * normal force
= muK * N
= 0.10 * 708 N
=70.8 N
Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.