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3 years ago

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My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin

es having length d and vertical height h and slide down. Rank the order, from greatest to smallest, of the final kinetic energies of the masses when they reach the bottom of the inclines after having traveled their full lengths. If any of the situations yield the same kinetic energies, give them the same ranking. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.) a. m 70 kg; h = 8 m; and d = 40 m b. m 70 kg; h 8 m; and d 20 m c. m 35 kg; h 4 m; and d 40 m d. m 7 kg; h 16 m; and d 20 m e. m-7 kg;h-4 m; and d=20 m f. m105 kg; h-6.0 m; and d 30.0 m

1 answer:

Tresset [83]3 years ago

8 0

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0 ⇒ K₀

and h₁ = 0 ⇒ U₁ = 0

we get

U₀ = K₁

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

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