A freight company uses a compressed spring to shoot 1.70kg packages up a 1.00m high frictionless ramp into a truck. The spring constant is 344N/m and the spring is compressed 33.0cm . What is the speed of the package when it reaches the truck?
Please explain! 🙂
also part B) A careless worker spills his soda on the ground just before the ramp. This creates a 50.0 cm long sticky spot with a coefficient of kinetic friction 0.300. Will the next package make it into the truck?
1 Answer
-
A) The potential energy in the spring
PS = ½kx²
This maximum energy available will be split between increase in potential energy and remaining kinetic energy
PE = mgh
KE = ½mv²
PS = PE + KE
KE = PS – PE
½mv² = ½kx² – mgh
v² = kx²/m – 2gh
v = √(kx²/m – 2gh)
v = √(344(0.33)²/1.7 – 2(9.81)(1.0))
v = 1.55 m/s
B) Now the maximum available energy will be split between the gain in potential energy, the loss to friction and the velocity remaining
PS = PE + Fd + KE
KE = PS – PE – Fd
½mv² = ½kx² – mgh – μmgd
v² = kx²/m – 2g(h + μd)
v =√(kx²/m – 2g(h + μd))
v =√(344(0.33²)/1.70 – 2(9.81)(1.00 + 0.300(0.500))
v =√-0.527
as this is an irrational number, the velocity goes to zero and the package does not make it to the top of the ramp
