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3 years ago

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A block of mass m, initially held at rest on a frictionless ramp a vertical distance H above the floor, slides down the ramp and

onto a floor where friction causes it to stop a distance D from the bottom of the ramp. The coefficient of kinetic friction between the box and the floor is ?k.
What is the macroscopic work done on the block by all forces during this process?

2 answers:

-Dominant- [34]3 years ago

4 0

Answer:

Explanation:

Check attachment for solution

bekas [8.4K]3 years ago

4 0

Answer and Explanation:

Total macroscopic workdone = Δ K.E

Where K.E = Kinetic Energy

Let V₁ = Initial speed of the block and

V₂ = Final speed of the block

Δ K.E = 1/2 MV₂² - 1/2 MV₁²

Δ K.E = 1/2M(V₂-V₁)²

Since the block was initially at rest, V₁ = 0

Since the block eventually stopped due to friction, V₂ = 0

Δ K.E = 1/2M * 0

Δ K.E = 0 Joules

N.B: The workdone by gravity is Mgh

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Explanation:

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Explanation:

In this question we have given

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resistivity of gold, $r=2.44\times 10^{-8} ohm.m$

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Temperature, $T= 20^oC$

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Here,

$A= \pi \times r^2$ ...........(3)

here A is the cross sectional area of cylendrical gold wire

From equation 2 and 3

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Put values of resistivity, A and L in equation 5, we got

$R = \frac{2.44 \times 10^{-8} ohm.m \times 2050 m}{2.53\times 10^{-11} m^2}$

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