Question

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.)

Answer 1

Potential energy is

`U=(1)/(2k)(mg(\mu_(s)\,cos\theta+sin\theta))^(2)\\`

Step-by-step explanation:

Given situation is shown in figure 1

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.

`F_(n)= mg\,cos \theta---(1)\\`

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

`T=F_(f)+mg\,sin\theta----(2)`

Frictional force is:

`F_(f)=\mu_(s)F_(n)---(3)`

From (1)

`F_(f)=\mu_(s)mg\,cos\theta`

(2) becomes then

`T=\mu_(s)mg\,cos\theta+mg\,sin\theta----(2)`

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

`T=kx---(4)`

Then (2) becomes

`kx=\mu_(s)mg\,cos\theta+mg\,sin\theta`

`x=(mg(\mu_(s)\,cos\theta+sin\theta))/(k)`

Potential energy is given as

`U=(1)/(2)kx^(2)\\\\U=(1)/(2)k((mg(\mu_(s)\,cos\theta+sin\theta))/(k))^(2)\\\\U=(1)/(2k)(mg(\mu_(s)\,cos\theta+sin\theta))^(2)\\`