hmm I can see what you are saying that the two forces can be the same, but I don't quite understand why the Work gravity does not affect it? since the lift is moving upwards, does gravity not affect it? sorry if its a stupid question. thanks two both of you guys for your help.
lowlypion there is no speed given in the question. if I work it out then would it be V = \sqrt{2K / m} , what value would I use for Kinetic energy? Wnet or W?
Homework Statement
a block of mass m = 20 kg is placed on the floor of a lift (mass M = 1000 kg), the lift is being pulled up distance d =25m by a cable, the normal force on the block FN has a constant magnitude of 300N, how much work is done on the lift by the force on the cable?Homework...
m = mass, b = damping constant, t = period ( not clear from my notes so I assumed was period), k = spring constant, xm2 = oscillation amplitude.
period t I used the forumla t = 2pi\sqrt{m/k} which gave me => 0.31s
for xm2 = oscillation amplitude I used t = ((-2 (m) ln 1/2) / b) = 4.27s...
o.k looking at my notes i can see that Mechanical energy E{t} = \frac{1}{2}kxm2 e-bt/m so using this formula I get: \frac{1}{2}E{t}= \frac{1}{2}kxm2 e-bt/m plugging in the values:
\frac{1}{2}E{t}= \frac{1}{2}80x4.27m2 e-0.065x0.31/0.2 = 659.30 / 2 => 329.65 J
and then rearranging the same...
Hi guys I am back again, I'm kind of stuck on part c of this question:
how long does it take for the inital mechanical energy of the damped system to halve its value?
part b asked how long it would take for the amplitude to halve, which i worked out to be 4.27s.
I have used 1/2 x m x Vm^2...
Homework Statement
Hi guys the question is: a mass spring-damper system is positioned between two rigid surfaces, if mass m = 200g, spring constant k = 80 Nm-1, and damping pot of coefficient 65 gs-1. The mass is pulled 5cm down from its equilibrium position and then released. What is the...