Question

The object with mass 0.5kg stationary at A is pull by a constant force 3N with a fixed ideal pulley at height 1m.The average coefficient of kinetic friction between the object and the ground is 0.3.
Find the velocity at B

Answer 1

http://openstudy.com/study#/updates/5250ff50e4b0bfb1fb4426c4

GRAPH AND INFORMATION

Answer 2

The answer is 1.7ms^-1

Btw, Thanks for helping,
Anyone explain?

Answer 3

Trying with work and energy. The increment in kinetic energy must be equal to the work done by the forces applied to the block,
\[\frac{1}{2}mv^2=\int\sum Fdx=\int (-F_r+T\cos\theta)dx\]Where,
\[F_r=\mu N=\mu(mg-T\sin\theta)\]and,
\[\tan\theta=1/x\Rightarrow x=1/\tan\theta\Rightarrow dx=-d\theta/\sin^2\theta \]So,
\[\frac{1}{2}mv^2=\left|\left[ -\frac{\mu m g}{\tan\theta}-T\ln\left(\tan(\theta/2)\right)+\frac{T}{\sin\theta }\right |\right]_{30}^{60}\]Solving for v,
\[v\approx 1.77\ m/s\]

Answer 4

Thanks!
I was thinking about energy-workdone also but don't know how to do the integral part ...

Answer 5

It was a little tricky, I think. I put the absolute value, in order to obtain a real velocity, but I think it is justified because the object is decreasing its distance to our origin. Anyways, I hope it helps.