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skelet666 [1.2K]
3 years ago
7

A 3.50 kg block is pushed along a horizontal floor by a force of magnitude 15.0 N at an angle θ = 30.0° with the horizontal. The

coefficient of kinetic friction between the block and the floor is 0.250.
Calculate the magnitudes of (a) the frictional force on the block from the floor and
(b) the block's acceleration.
Physics
1 answer:
Crank3 years ago
8 0

Answer :

The frictional force on the block from the floor and the  block's acceleration are 10.45 N and 0.73 m/s².

Explanation :

Given that,

Mass of block = 3.50

Angle = 30°

Force = 15.0 N

Coefficient of kinetic friction = 0.250

We need to calculate the frictional force

Using formula of frictional force

F_{k}=\mu N

F_{k}=\mu (F\sin\theta+mg)

F_{k}=0.250\times(15\times\sin30^{\circ}+3.50\times9.8)

F_{k}=0.250\times41.8

F_{k}=10.45\ N

(II). We need to calculate the block's acceleration

Using newton's second law of motion

F=ma

a=\dfrac{F}{m}

a=\dfrac{F\cos\theta-F_{k}}{m}

a=\dfrac{15.0\cos30^{\circ}-10.45}{3.50}

a=0.73\ m/s^2

Hence, The frictional force on the block from the floor and the  block's acceleration are 10.45 N and 0.73 m/s².

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Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

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This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

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4 0
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In the lab downstairs physics majors use a rotating mirror to measure the speed of light within a few percent of the actual valu
iris [78.8K]

The number of complete cycles the rotating mirror goes through before the angular velocity gets to zero is approximately 1166.8 revs

<h3>What is angular velocity?</h3>

Angular velocity is the ratio of the angle turned to the time taken.

The kinematic equation for angular velocity are presented as follows;

ω = ω₀ + α·t

θ = θ₀ + ω₀·t + 0.5·α·t²

Where;

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ω₀ = The initial angular velocity of the mirrors = 115 rad/s clockwise

α = The angular acceleration = (115  - (-115))rad/s/(85 s) = -46/17 m/s²

t = The duration of the motion;

When the angular velocity, ω is zero, we get;

0 = 115 - 46/17·t

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Which indicates;

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θ = 7331.25/(2×π) ≈ 1166.8 rev

The mirrors would have turned through approximately 1166.8 revolutions when the angular gets to zero

Learn more about angular velocity and acceleration here:

brainly.com/question/13014974

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