Rianar9ockmorevanPa
contestada

An object of mass m is at rest at the top of a smooth slope of height h and length L. The coefficient of kinetic friction between the object and the surface, μk, is small enough that the object will slide down the slope if given a very small push to get it started. Find an expression for the object's speed at the bottom of the slope.
Express your answer in terms of the variables m, μk, L, h, and appropriate constants.

Respuesta :

saadhussain514

Answer:

The expression is

v = √(2gh - 2µk*gLcos(arcsin(h/L))

Explanation:

The expression "very small push" means that the initial Kinetic energy KE = 0.

So we have

Net Kinetic Enrgy KE = initial PE - work done

½mv² = mgh - µk*mgLcos(arcsin(h/L))

Multiplying it by 2/m

v² = 2gh - 2µk*mgLcos(arcsin(h/L))

Now taking the square root

v = √(2gh - 2µk*gLcos(arcsin(h/L))

This is the final expression

Answer:

speed at the end is given as

[tex]v = \sqrt{(2gh - 2\mu g\sqrt{L^2 - h^2})}[/tex]

Explanation:

As we know that here we can find the final speed of object by using work energy theorem

here when object slides down the inclined plane there will be two forces that will work on it

1) Work done by gravity

2) Work done by friction

Now we have work done by gravity

[tex]W_g = F_g \times h[/tex]

here we know that

[tex]F_g = mg[/tex]

now work done by friction force

[tex]W_f = -F_f \times L[/tex]

now here we know that friction force is given as

[tex]F_f = \mu mgcos\theta[/tex]

here we know that

[tex]cos\theta = \frac{\sqrt{L^2 - h^2}}{L}[/tex]

[tex]W_f = -\mu mgL (\frac{\sqrt{L^2 - h^2}}{L})[/tex]

now we have

[tex]mgh - \mu mgL (\frac{\sqrt{L^2 - h^2}}{L}) = \frac{1}{2}mv^2[/tex]

now the final speed is given as

[tex]v = \sqrt{(2gh - 2\mu g\sqrt{L^2 - h^2})}[/tex]

so above is the speed by which it will slide at the end of inclined plane

Otras preguntas