A smooth body is released from rest A at a point A at the top of a smooth curved track of vertical height 40 cm, what is the speed of the body at d bottom of the curved track. How far along the adjoining smooth inclined plane will d body go?
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26
Answer:
Explanation:
=> In the absence of friction, the energy of the body is conserved so that on each point of the track we have:
mv²/2 + mgH = constant
0 + mgh = 1/2 mv² + mg
1/2 mv² = mg (40 - 1)
v² = 39 * g * 2
v = √76440
v = 276.4 cm/s
v = 2.8 m/s
Now, v² = u² + 2gs
v² = u² + 2g( sin 30°) s
Final velocity will be zero here
2.8 * 2.8 = -2 * 9.8 * s / 2
=> neglect negative sign, sice distance can not be negative.
s = 0.8m = 80cm
Answered by
2
Answer:
Answer is 2.8 m/s, 80cm
I hope it helps you.
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