Mass and Kinetic Energy Essay

964 Words Apr 16th, 2013 4 Pages
Disk With Weight:

A 15 kg uniform disk of radius R = 0.25 m has a string wrapped around it, and a m = 3 kg weight is hanging on the string. The system of the weight and disk is released from rest.

a) When the 3 kg weight is moving with a speed of 2.2 m/s, what is the kinetic energy of the entire system?
KETOT = KEwheel+KEweight
= (1/2)(I)(w2)+(1/2)(m*v2)
=(0.5* v2)(m+1/2M)
=0.5*(2.2^2)*(3+(.5*15)) J

b) If the system started from rest, how far has the weight fallen?
H = KETOT/MG
= 0.5*(2.2^2)*(3+(.5*15))/(3*9.8) m


c) What is the angular acceleration at this point?
Remember that a = αR, or α = a/R


Solve for acceleration by using vf2=vi2+2ax
…show more content…
Now, what is the translational kinetic energy of the sphere at the bottom of the incline?
=mgh (because there’s no KErot, this is what ends up happening)
4.2*3.7sin(33)*9.8




Bar and Weights

A beam of mass mb = 10.0 kg, is suspended from the ceiling by a single rope. It has a mass of m2 = 40.0 kg attached at one end and an unknown mass m1 attached at the other. The beam has a length of L = 3 m, it is in static equilibrium, and it is horizontal, as shown in the figure above. The tension in the rope is T = 637 N.

a) Determine the unknown mass m1, at the left end of the beam.
Sum of the forces = 0 because the system is in equilibrium, so we can do F= ma, ma= 0, so F=0
m1g+m2g+m3g-Ft= 0
g(m1+m2+m3)=Ft
m1= (Ft/g)-(m2+m3)
15


b) Determine the distance, x, from the left end of the beam to the point where the rope is attached. Note: take the torque about the left end of the beam.
=2.07




Hanging Sign

A sign has a mass of 1050 kg, a height h = 1 m, and a width W = 4 m. It is held by a light rod of length 5 m that is perpendicular to a rough wall. A guy wire at 23° to the horizontal holds the sign to the wall. Note that the distance from the left edge of the sign to the wall is 1 m.…

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