If m 1 falls down 2.5 m (d) What is the total change in kinetic energy ? K syste

Question

If m1 falls down 2.5 m

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

(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 hcylinder 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?

Explanation / Answer

(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

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