This is going to be very long question.
A 1 kg block situated on a rough incline is connected to a spring constant 100Nm as shown in fig. The block is released from rest with the spring in the unstretched position. The block moves 10cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume the spring has a negligible mass and the pulley is frictionless.
The way I tackled the question was by using forces acting at equilibrium. The forces acting on the block are: 1.The force due spring up the incline. 2.The force due friction up the incline . 3.Components of gravity.
$R=mg \cos 37$.
The force of friction $f$ will be up the incline $f=\mu mg\cos 37$.
The component of gravity along the downward slope = $mg \sin 37$.
Putting it all together .
$mg\sin 37 - kx - \mu mg \cos 37 = 0$.
$mg \sin 37-kx = \mu mg \cos 37$.
Here this is get messed up, $\mu$ turns out to negative once you put the values in. If we change the direction of frictional force, the answer comes out to be 0.5. But the real solution floating on the internet solves the question by using work done .
Work done by friction $-\mu xmg \cos 37$. Work done by gravity $xmg \sin 37$.
Energy stored $\frac{kx^2}{2}$.
Therefore $xmg \sin 37 - \mu xmg \cos 37 = \frac{kx^2}{2}$.
Putting values gives the answer around 0.125.
Where did I went wrong?