Question

Block A has a weight of 300N and block B has a weight of 50 N. If the coefficient of kinetic friction between the incline and block A is ${\mu }_k$ = 0.2. Determine the speed of block A after it moves 1m down the plane, starting from rest. Neglect the mass of the cord and pulleys.

Answer 1

R.E.F image

$N=300cos\theta$

$=300\times \frac{4}{5}=240N$

Let us take,

Velocity of block be A after it moves 1 m

down the plane be $V$

and velocity of block B be $V_B$

We know that,

in pulley system

T.U = constant

$\Rightarrow 27.u=\pi u_B$

$\Rightarrow u_B-2u$

and $T.x=$ constant

$\Rightarrow 27\times 1=f_2$

$x=2m$

Since $N=240N$

$\Rightarrow f=uN=0.2\times 240$

$f=48N$

Applying energy conservation

Decrease in potential energy of block A

= Increase in kinetic energy of block A

+ Increase in kinetic energy of block B

+ Increase in potential energy of block B

+ work done by friction force

$\Rightarrow 300\times 1sin\theta =\frac{1}{2}\times \frac{300}{g}\times v^2+\frac{1}{2}\times \frac{50}{g}(2v{)}^2+50\times 2+48\times 1$

$300\times \frac{3}{5}=15v^2+10v^2+100+48$

$=180=25v^2+148$

$v^2=\frac{180-148}{25}$

$v^2=\frac{32}{25}$

$v=\frac{4\sqrt{2}}{5}$