PHYSICS QUESTION: WORK AND ENERGY?

A bicyclist travels up a 14.1-m hill and reaches the top with a speed of 9.7 m/sec. At that point, the bicyclist stops pedaling and coasts down the hill and up the next one. If the next hill is 2.3 meters high how fast will the bicyclist be traveling when it reaches the top of the second hill?

1 Answer

Relevance
  • 1 year ago
    Favourite answer

    energy is conserved.

    initial energy = KE + PE = ½mV² + mgh

    = m(½V² + gh) = m(½9.7² + 9.8•14.1)

    final energy = m(½V² + 9.8•2.3)

    set them equal and solve for V

    ½9.7² + 9.8•14.1 = ½V² + 9.8•2.3

    ½V² = ½9.7² + 9.8•14.1 – 9.8•2.3

    ½V² = ½9.7² + 9.8(14.1 – 2.3)

    ½V² = ½9.7² + 9.8(11.8)

    V² =9.7² + 19.6(11.8)

    V = √(9.7² + 19.6•11.8)

    you have a calculator...

    Kinetic Energy in J if m is in kg and V is in m/s

    KE = ½mV²

    Potential Energy in J

    PE = mgh

    g is the acceleration of gravity (9.8 m/s² on surface of earth)

    • Yolanda1 year agoReport

      thank you so much!!

    • Commenter avatarLog in to reply to the answers
Still have questions? Get answers by asking now.