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Ganezh [65]
3 years ago
12

This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag

nitude of the force is different in the two cases, while the directional angle θ is the same. Kinetic friction exists between the block and the wall, and the coefficient of kinetic friction is 0.340. The weight of the block is 51.0 N, and the directional angle for the force is θ = 44.0°. Determine the magnitude of when the block slides (a) up the wall and (b) down the wall.
Physics
1 answer:
sweet-ann [11.9K]3 years ago
3 0

Answer:

Explanation:

Given

coefficient of kinetic friction(\mu _k)=0.34

inclination \theta =44

weight of block=51 N

(a) When block is moving upward friction force acts downward

thus

Fsin\theta -W-f_r=0

as block is moving with constant velocity thus F_{net} is zero

f_r=\mu _kN=0.34\times Fcos\theta

F\left ( \sin \theta -\mu \cos \theta \right )=W

F=\frac{51}{0.45}=113.31 N

(b)When Block slides down the wall friction changes its direction to oppose the block

Fsin\theta -W+f_r=0

F\left ( \sin \theta +\mu \cos \theta \right )=W

F=\frac{W}{\left ( \sin \theta +\mu \cos \theta \right )}

F=\frac{51}{0.939}=54.299 N

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Ierofanga [76]

By Newton's second law, the net vertical force acting on the object is 0, so that

<em>n</em> - <em>w</em> = 0

where <em>n</em> = magnitude of the normal force of the surface pushing up on the object, and <em>w</em> = weight of the object. Hence <em>n</em> = <em>w</em> = <em>mg</em> = 196 N, where <em>m</em> = 20 kg and <em>g</em> = 9.80 m/s².

The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of <u>static</u> friction <em>µ</em> is such that

80 N = <em>µ</em> (196 N)   →   <em>µ</em> = (80 N)/(196 N) ≈ 0.408

Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of <u>kinetic</u> friction <em>ν</em> is

40 N = <em>ν</em> (196 N)   →   <em>ν</em> = (40 N)/(196 N) ≈ 0.204

And so the closest answer is C.

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3 0
3 years ago
A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte
vaieri [72.5K]

Answer:

Part a)

H = 44.1 m

Part b)

y = 13.48 m

Part c)

d = 8.86 m

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

t = \frac{v sin\theta}{g}

now we have

3 = \frac{vsin\theta}{9.8}

v sin\theta = 29.4 m/s

Now maximum height above ground is given as

H = \frac{v^2sin^2\theta}{2g}

H = \frac{29.4^2}{2(9.8)}

H = 44.1 m

Part b)

Height of the fence is given as

y = (vsin\theta) t - \frac{1}{2}gt^2

y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)

y = 13.48 m

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

x = v_x t

97.5 = v_x (5.5)

v_x = 17.72 m/s

now total time of flight is given as

T = 3 + 3 = 6 s

so range is given as

R = v_x T

R = (17.72)(6)

R = 106.4 m

so the distance from the fence is given as

d = 106.4 - 97.5

d = 8.86 m

7 0
3 years ago
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