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deff fn [24]
3 years ago
8

A uniform meter stick of mass $M$ has a half-filled can of fruit juice of mass $m$ attached to one end. The meter stick and the

can balance at a point 21.2 cm from the end of the stick where the can is attached. When the balanced stick-can system is suspended from a scale, the reading on the scale is 2.58 N. Calculate the mass of the meter stick.
Physics
1 answer:
notka56 [123]3 years ago
7 0

Answer:

111.6 g

Explanation:

Given

m + M = 2.58 / 9.8

= 0.2632 kg

When the can of fruit juice is balance from scale , we get following relation

M x ( 50 - 21.2 ) = m x 21.2 ( balancing the torque due to weight of scale and can about the balancing point )

M x 28.8 = 21.2 m

= 21.2 ( 0.2632 - M )

= 5.58 - 21.2 M

M ( 28.8 + 21.2 ) = 5.58

M = .1116 kg

= 111.6 g

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Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

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from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + 0.5mV1^{2} = mgh2 + 0.5mV2^{2}

where

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  • speed at lowest point = V1
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  • at the lowest point the speed is maximum (V2 is the maximum speed)
  • therefore the equation becomes mgh1 + 0.5mV1^{2} = mgh2

      m(gh1 + 0.5V1^{2}) = m(gh2)

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