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nikdorinn [45]
3 years ago
15

A mass M of 3.80E-1 kg slides inside a hoop of radius R=1.10 m with negligible friction. When M is at the top, it has a speed of

5.35 m/s. Calculate the size of the force with which the M pushes on the hoop when M is at an angle of 34.0°.
Physics
1 answer:
Vikentia [17]3 years ago
3 0

Answer:

N = 26.59 N

Explanation:

given,

mass = 0.38 kg

radius of the hoop = 1.10 m

speed = 5.35 m/s

force = ?

now,

\dfrac{1}{2}mv_t^2 + mg(2R) = \dfrac{1}{2}mv^2 + mgR(1-cos \theta)

mv^2 = mv_t^2 + 2mgR(1 + cos \theta)

we know that,

N - mgcos \theta = \dfrac{mv^2}{R}

N - mgcos \theta = \dfrac{mv_t^2 + 2mgR(1 + cos \theta)}{R}

N - mgcos \theta = \dfrac{mv_t^2 }{R}+ 2mg(1 + cos \theta)

N  = \dfrac{mv_t^2 }{R}+ 2mg + 3mgcos \theta)

N  = \dfrac{0.38\times 5.35^2 }{1.1}+ 2\times 0.38\times 9.8 + 3\times 0.38 \times 9.8 cos 34^0)

N = 26.59 N

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<u>Answer:</u>

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