Question

If m₁ falls down 2.5 m

(d) What is the total change in kinetic energy ?K_system of the system. Note: A positive answer indicates an increase in kinetic energy and a negative answer indicates a decrease in kinetic energy.
?K_system =

(e) What is the final velocity v of the system?
v =

A group of students perform the same "Conservation of Mechanical Energy" experiment that you performed in lab by allowing a solid sphere and then a solid cylinder to roll down the ramp. The solid sphere was released from a height of 15.3 cm. From what height h_cylinder should the solid cylinder be released so that it has the same speed as the solid sphere when it reaches the bottom of the ramp?

Answer 1

(1)

d) After falling 2.5m, what is the velocity of m1?
To find m1's velocity after falling 2.5m
Use Vf^2=Vi^2+2ad => Vf = sqrt(2*9.8*2.5) = 7m/s
Assuming the inclined plane is frictionless
Since m2 is roped to m1, it's velocity up the ramp will be 7m/s too
The total increase in KE is then 0.5[7+5]*7^2 = 294J
e) 7m/s assuming m1 is free to fall. 7m/s is the velocity after falling 2.5m

(2)

mgh = 1/2mv^2 + 1/2 . 2/5 mv^2 (for the sphere )

0.153*9.8 = v^2 * 0.7

v = 0.1464 m/s

For cylinder

2gh = 14.64^2 + 0.5 * 14.64^2

h = 0.0016403 m = 16.403 cm