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?
- billrussell42Lv 71 year agoFavourite 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)