A block of mass m rests on an inclined plane. The coefficient of static friction between the block and the plane is μs. A gradually increasing force is pulling down on the spring (force constant k). Find the potential energy U of the spring (in terms of the given symbols) at the moment the block begins to move. (Use any variable or symbol stated above along with the following as necessary: g and θ.)

In order to calculate the potential energy at the moment block started moving we first have to calculate the net force acting on the block. For this consider figure 2

Forces acting in vertical direction:

Normal force Fn acting on block is in upward direction which is balanced by horizontal component of weight of block acting in opposite direction.

[tex]F_{n}= mg\,cos \theta---(1)\\[/tex]

Forces acting in horizontal direction:

Tension in string is equal and opposite to frictional force and and vertical component of weight as show in figure 2

[tex]T=F_{f}+mg\,sin\theta----(2)[/tex]

Frictional force is:

[tex]F_{f}=\mu_{s}F_{n}---(3)[/tex]

From (1)

[tex]F_{f}=\mu_{s}mg\,cos\theta[/tex]

(2) becomes then

[tex]T=\mu_{s}mg\,cos\theta+mg\,sin\theta----(2)[/tex]

If spring is not accelerating then tension in rope must be equal spring force i.e

[tex]T=kx---(4)[/tex]

Then (2) becomes

[tex]kx=\mu_{s}mg\,cos\theta+mg\,sin\theta[/tex]

[tex]x=\frac{mg(\mu_{s}\,cos\theta+sin\theta)}{k}[/tex]

Potential energy is given as

[tex]U=\frac{1}{2}kx^{2}\\\\U=\frac{1}{2}k(\frac{mg(\mu_{s}\,cos\theta+sin\theta)}{k})^{2}\\\\U=\frac{1}{2k}(mg(\mu_{s}\,cos\theta+sin\theta))^{2}\\[/tex]