A spring with a 900 N/m spring constant is used to launch a 3 kg box up a curved

Question

A spring with a 900 N/m spring constant is used to launch a 3 kg box up a curved ramp. The spring has a relaxed length of 60 cm and is compressed to a new length of 20 cm. Between the spring and the ramp is a 10 meter long flat surface with a 0.15 coefficient of kinetic friction. The curved ramp is frictionless. Our goal is to determine the maximum height the box will reach on the ramp.

a) Define your system

b) Define your initial and final moments for conservation of energy.

c) Write your conservation of energy equation and solve for the maximum height.

d) In words, why is the maximum height not 2.4 meters? Where is the “missing” energy?

e) Without doing any calculations, when the box slides back down and runs into the spring, will the spring compress more, less, or the same amount as it was originally compressed? Why?

f) Use conservation of energy to calculate how far the spring will compress when struck by the box.

Explanation / Answer

b) Initial energy = elastic potential energy = 1/2 kx^2 = 1/2 (900) (0.2)^2=18 J

energy imparted to 3kg box=18 J

Initial velocity of 3kg box 1/2 kx^2 = 1/2 mv^2

v^2= 36 /3= 12

Velocity after travelling a distanc e= 10 m = V^ 2= 12 - 2 (10) (0.15 x 9,8) = -17.4 which shows that 3 kg nlock will not be able to cross the 10 m flat road. It will stop in mid of the flat road ...

Max distance travelled by cup = 2(0.15) ( 9.8) (x) = 12

x= 4.08 m ( after travelling 4.08 m the cup will stop on the flat road)

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