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A 100 kg block slides down a frictionless ramp which is at an angle of 26.6 ∘ with the horizontal. Find the magnitude of the velocity of the block when it gets to the bottom of the ramp if the length of the incline is 50 meters. V=_____m/s

Sagot :

LammettHash

The block accelerates down the ramp due to the component of its weight that acts parallel to the ramp; this force has magnitude

(100 kg) g sin(26.6°) ≈ 438 N

(Note that the positive sign here means we take "down the ramp" to be the positive direction.)

Since the ramp is frictionless, this is the only force acting on the block in this direction. By Newton's second law, the block's acceleration is a such that

438 N ≈ (100 kg) a   ⇒   a ≈ 4.38 m/s²

The block accelerates uniformly, so that it attains a speed v as it moves 50 m such that

v² = 2a (50 m)

Solve for v :

v = √(2a (50 m)) ≈ 20.9 m/s

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