Physics
13.06.2023 01:29
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A small block with mass 0.0425 kg slides in a vertical circle of radius 0.525 m

A small block with mass 0.0425 kg slides in a vertical circle of radius 0.525 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force exerted on the block by the track has magnitude 4.05 N. In this same revolution, when the block reaches the top of its path, point B, the magnitude of the normal force exerted on the block has magnitude 0.660 N. Required:
How much work was done on the block by friction during the motion of the block from point A to point B?
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ninmagic
ninmagic
4,9(33 marks)

W=-0.501075J \approx -0.5J

Explanation:

From the question we are told that

Mass of block M=0.425kg

Radius of circle r=0.525m

Force  of track F_t=4.05N

Normal Force on blockF_n=0.660

Generally the equation for Velocity of block is mathematically Given by the law of conservation of energy

4.05=(0.425kg*9.8)+(0.425kg*v^2)/0.525

v=\sqrt{\frac{0.525*(4.05-(0.0425*9.8)}{0.0425} }

V=-6.69 \approx 7m/s

For initial velocity

0.660 = 0.0425 x u^2/0.525 - 0.0425 x 9.8

v=\sqrt{0.525*\frac{(0.660+(0.0425*9.8)}{0.0425} }

V=-3.646 \approx 4m/s

Generally the equation for work done is mathematically Given by

Wf=(0.5mu^2 - 0.5mv^2) -mg(2R)

Wf=(0.5(0.0425)(4) - 0.5(0.0425)(7) -(0.0425)(9.8)(2*0.525)

W=-0.501075J \approx -0.5J

jchavez5601
jchavez5601
5,0(96 marks)

Step-by-step explanation:

f(4) = 42 + 5 =21, f(-10) = (-10)2 +5 = 105 or alternatively f: x → x2 + 5.

The phrase "y is a function of x" means that the value of y depends upon the value of x, so:

y can be written in terms of x (e.g. y = 3x ).

If f(x) = 3x, and y is a function of x (i.e. y = f(x) ), then the value of y when x is 4 is f(4), which is found by replacing x"s by 4"s .

If f(x) = 3x + 4, find f(5) and f(x + 1).

f(5) = 3(5) + 4 = 19

f(x + 1) = 3(x + 1) + 4 = 3x + 7

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