Question

Two clever kids from Kentucky use a huge spring with spring constant k= 890 N/m to launch their sled at the top of a 9.5 m high hill. The mass of kids plus the sled is 80 kg. The brilliant kids are able to compress the spring 2.6 m.

a. Determine their speed at the bottom of the hill. Neglect friction. Use conservation of energy.

b. What fraction of their final kinetic energy was initially stored in the spring?

c. Now assume that the surface of the snow is not frictionless and has a coeficcient of kinetic friction uk=0.100. Determine their speed at the bottom of the hill. Assume that the slope has an average incline of 60 degree.

Please show your work clearly and write step by step solution including numeric substitutions and etc. To make it easier, please do this in paper and include the pictures. thanks

Answer 1

k= 890 N/m
Height h = 9.5 m
mass m = 80 kg
x compression = 2.6 m

(a)
Intial Energy = Spring Energy + Potential Energy
Final Energy = Kinetic Energy

Using conservation of energy
Intial Energy = Final Energy
Spring Energy + Potential Energy = Kinetic Energy
0.5*kx^2 + m*g*h = 0.5*m*v^2
0.5*890*2.6^2 + 80*9.8*9.5 = 0.5*80*v^2
v = 16.17 m/s
Speed at the bottom of the hill, v = 16.17 m/s

(b)
Final Kinetic Energy =  0.5*m*v^2 = 0.5*80*16.17^2 = 10458.7 J
Initial Spring Energy = 0.5*890*2.6^2 = 3008.2

Fraction of their final kinetic energy was initially stored in the spring = 3008.2/10458.7 = 0.287

(c)
Intial Energy = Spring Energy + Potential Energy
Final Energy = Kinetic Energy + Energy lost in friction

Using conservation of energy
Intial Energy = Final Energy
Spring Energy + Potential Energy = Kinetic Energy + Energy lost in Friction
0.5*kx^2 + m*g*h = 0.5*m*v^2 + u*m*g*cos(60)* (9.5/sin(60))
0.5*890*2.6^2 + 80*9.8*9.5 = 0.5*80*v^2 + 0.1*80*9.8*cos(60) * (9.5/sin(60))
v = 15.83 m/s
Speed at the bottom of the hill, v = 15.83 m/s