Physics Question about spring compression? Grade12?

NetherCraft 0

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! 🙂

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

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