A block is released from rest, at a height h, and allowed to slide down an inclined plane. There is friction on the plane. At the bottom of the plane, there is a spring that the block will compress. After compressing the spring, the block will slide up the plane to some maximum height, hA, after which it will again slide back down. How much work is done on the block between its release at height h and its ascent to its next maximum height?

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moya1316 moya1316 · Answer 1 · 2019-11-12T11:19:34+01:00

Answer:

[tex]W_{fr}[/tex] = - mg (h - hₐ)

Explanation:

For this exercise we must use the energy work relationship

W = ΔEm

Where work is the work of the friction forces along the plane as the force of friction opposes the movement is always negative

W = -fr d

The energy can be searched for two points of interest, when it is going down, let's use the highest point and the lowest point

Initial, highest point

Em₀ = U = m g and

Final, lower at maximum spring compression

[tex]Em_{f}[/tex]= K = ½ m v²

Let's replace

-fr₁ d₁ = ½ m v² - m g y

We do the same for when the plane bounces and goes up

Initial, lower

Em₂ = K = ½ mv²

Higher

Em₃ = U = m g ya

-fr2 d2 = mg yₐ - ½ mv²

Let's write our system of equations

-fr₁ d₁ = ½ m v² - m g y

-fr₂ d₂ = mg yₐ - ½ mv²

Let's add

-fr₁d₁ –fr₂d₂ = mg yₐ - mg y

[tex]W_{fr}[/tex] = mg (yₐ -y)

[tex]W_{fr}[/tex] = - mg (h - hₐ)