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rjkz [21]
3 years ago
7

A 0.50-$kg$ block slides along a small track with elevated ends and a flat central part. The flat part has a length L = 1.00 $m$

. The curved portions of the track are frictionless, but for the flat part the coefficient of kinetic friction is 0.144. The block is released from rest from a height h = 70 $cm$ on the left curved portion of the track. Calculate the maximum height reached by the block on the right curved portion of the track.

Physics
1 answer:
Annette [7]3 years ago
6 0

Answer:

y=55.6 cm

Explanation:

Given that

m=0.5 kg

L= 1 m

h= 70 cm = 0.7 m

coefficient of kinetic friction( μ ) = 0.144

Lets take maximum of height achieved by block on the right side

The friction force on the flat part

fr= μ  m g

Initial velocity at the top left is zero and the final speed of the bock at right top will be zero .

we know that work done by all force = Change in kinetic energy

Here initial and final speed is zero so the change in kinetic energy is zero.

m g h - fr . L - m g y=0

fr= μ  m g

m g h - μ  m g .L  - m g y=0

h - μ.L - y = 0

By putting the values

0.7 - 0.144 x 1 - y = 0

y=0.556 m

y=55.6 cm  

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\gamma_{unk} = \frac{P_{bot} - \gamma_{oil}h_1}{h_2} = \frac{65\frac{kN}{m^2} - 8.5\frac{kN}{m^3}\times 5m}{1.5m} = \mathbf{15 \frac{kN}{m^3}}.

Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.

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what is the net force on an object that is experiencing a force of 25 N north, a force of 25 N south, a force of 50 N to the eas
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Answer:

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Answer:

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