http://www.ck12.org Chapter 6. Work and Energy
FIGURE 6.20
Consider trying to solve this problem using Newton’s Laws. What acceleration would you use along the ramp?
Since the ramp is curved, the acceleration is changing at every instant. This problem would be very difficult to solve
using Newton’s Laws and more information would be required as well. We would need to know the function that
describes the shape of the ramp and then we would need to make use of calculus, which is beyond the scope of this
book.
Check Your Understanding
- If the sled had fallen straight down from a height of 4.00 m, what speed would it have upon reaching the ground?
Answer:If you were to use the kinematic equation:vf^2 =vi^2 + 2 a∆x
wherea=− 9. (^81) sm 2 ,vi= 0 ,and∆x=− 4. 00 m
vf^2 = 0 + 2 (− 9. 81 )(− 4. 00 ) = 78. 4 ,vf= 8. 86 m/s
The speed at the bottom of the ramp is identical to the speed the sled would have had, had it fallen straight down.
Energy conservation is independent of the path the object takes, as long as there are no non-conservative forces
present.
Illustrative Example 4
InFigure6.21, an amusement park ride is constructed such that a car carrying passengers is initially locked against
a huge compressed spring. When the spring is released, the car is projected along a straight horizontal segment of
track before encountering a curved ramp. The system is assumed to have negligible friction.
What is the maximum height the car meets above the ground if the spring is compressed 0.75 m, the spring constant
is 25,000 N/m, and the car and passengers have a mass of 500 kg?
Answer:The initial energy is contained in the compressed spring which upon release will be communicated to the
car. After the car is free of the spring it will have its maximum kinetic energy which remains constant along the
horizontal until the car begins its rise along the curved ramp. At the height,h, all of the kinetic energy has been
transformed into potential energy.