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liraira [26]
4 years ago
12

A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet

ween the incline and the block is µk = 0.3
i) Determine the acceleration of the block.
ii) The block starts at rest from a height of 8m above the ground. What is the speed of the block when it reaches the bottom of the incline.
Physics
1 answer:
astraxan [27]4 years ago
5 0

Answer:

Acceleration = 2.35 m/s^{2}

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = \frac{mg}{2} = 0.5 mg

Friction along incline = umg cos30° = mg 0.3\times \frac{\sqrt{3}}{2}

Friction along incline  = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = \frac{net force}{mass} = 0.24 g = 2.35 m/s^{2}

The height of incline = 8 m

Length of the inclined edge = 16 m

v^{2}=u^{2}+2as

v^{2}= 2\times 0.24 \times 9.8\times 16

v= 8.67 m/s

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Four weightlifters (A-D) enter a competition. The mass, distance, and time of their lifts are shown in the table.
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Let Pa, Pb, Pc, and Pd be the powers delivered by weightlifters A, B, C, and D, respectively.

Use this equation to determine each power value:

P = W÷Δt

P is the power, W is the work done by the weightlifter, and Δt is the elapsed time.

A) Determining Pa:

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The weightlifter does work to lift the weight up a certain distance. Therefore the work done is equal to the weight's gain in gravitational potential energy. The equation for gravitational PE is

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PE is the potential energy, m is the mass of the weight, g is the acceleration of objects due to earth's gravity, and h is the distance the weight was lifted.

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B) Determining Pb:

Let us use our new equation derived in part A to solve for Pb:

Pb = mgh÷Δt

Given values:

m = 150.0kg

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h = 1.76m

Δt = 0.052s

Plug in the values and solve for Pb

Pb = 150.0×9.81×1.76÷0.052

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Pc = mgh÷Δt

Given values:

m = 200.0kg

g = 9.81m/s²

h = 1.50m

Δt = 0.217s

Plug in the values and solve for Pc

Pc = 200.0×9.81×1.50÷0.217

<u>Pc = 13600W</u>

D) Determining Pd:

Pd = mgh÷Δt

Given values:

m = 250.0kg

g = 9.81m/s²

h = 1.25m

Δt = 0.206s

Plug in the values and solve for Pd

Pd = 250.0×9.81×1.25÷0.206

<u>Pd = 14900W</u>

Compare the following power values:

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Pc is the lowest value.

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Answer:

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r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk

Velocity is the derivative of position, r:

V = dr/dt = 3.63 - 11.46t^j

A. x component of velocity, Vx = 3.63 m/s

B. y component of velocity, Vy = -11.46t

t = 3.99 secs,

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|V| = √(2091.2329 + 13.1769)

|V| = √(2104.4098)

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D. Angle of the velocity relative to the x axis, θ is given as:

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Answer:

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Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

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b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

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r= corresponding radial distance of each bug

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The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

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Option "e" is correct.

7 0
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