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PilotLPTM [1.2K]
3 years ago
15

A 60-kilogram sled is coasting with a constant velocity of 10 m/s over smooth ice. It enters a rough stretch of ice 6.0 m longs

in which the force of friction is 120 N. a) What is the acceleration during this stretch? B) With what speed dies it emerge from the rough patch?
Physics
1 answer:
klemol [59]3 years ago
6 0

Answer:

-2 m/s²

8.7 m/s

Explanation:

Using Newton's second law, the sum of forces in the x direction:

∑F = ma

-120 N = (60 kg) a

a = -2 m/s²

Given:

Δx = 6.0 m

v₀ = 10 m/s

a = -2 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (10 m/s)² + 2 (-2 m/s²) (6.0 m)

v = 8.7 m/s

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A block of mass m is set sliding up along a frictionless ramp inclined at angle θ relative to horizontal. The angle of the ramp
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1) h is constant

2) h is constant

3) h increases

Explanation:

"What happens to the height h as the angle θ is increased?"

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½ mv² = mgh

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h = v² / (2g)

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"What happens to the height h if θ and v are kept the same, but the mass m of the block increases?"

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"If we keep m and v the same, but now we add a small amount of kinetic friction to the ramp surface, how does the height h change as θ increases?"

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½ mv² = Fd + mgh

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The friction force can be found by drawing a free body diagram of the block.  There are three forces acting on the block.  Normal force perpendicular to the ramp, weight force pulling down, and friction force pushing down the ramp.

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F = Nμ

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Substituting:

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½ mv² = mgμ h / tan θ + mgh

½ v² = gμ h / tan θ + gh

½ v² = (gμ / tan θ + g) h

As θ increases, tan θ increases.  That makes gμ / tan θ + g decrease.  Since v is constant, h must increase.

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