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Sav [38]
3 years ago
13

A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon

tally on it. The coefficient of kinetic friction between the crate and floor is 0.26.
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?

Physics
2 answers:
Debora [2.8K]3 years ago
8 0

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force F_p by the worker must be equal to the friction force F_f on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N

b. The work is done on the crate by this force is the product of its force F_p and the distance traveled s = 4.35

W_p = F_ps = 79.1*4.35 = 344 J

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

W_f = F_fs = -79.1*4.35 = -344 J

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

W_p + W_f = 344 - 344 = 0 J

ycow [4]3 years ago
8 0

Answer:

(A) 79N

(B) W = 344J

(C) Wf= -344J

(D) W = 0J

(E) W = 0J

Explanation:

Please see attachment below.

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r = 8.7 m at 135 degree with positive X axis

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r = 8.7 cos135\hat i + 8.7 sin135\hat j

r = -6.15 \hat i + 6.15 \hat j

now we know that

r = r_1 + r_2

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r_2 = r - r_1

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tan\theta = \frac{-4.75}{-7.68}

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8 0
3 years ago
Select three different examples of accelerated motion. a body traveling in a straight line and increasing in speed a body travel
IrinaVladis [17]
This is the same question that I just answered.

Have present the definition of acceleration:

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A decrease in speed is a change in velocity, so it means acceleration.

3) a body traveling in a straight line at constant speed: FALSE.

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4 years ago
A person eats a container of strawberry yogurt. The Nutritional Facts label states that it contains 255 Calories (1 Calorie = 41
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Answer:

mass = 0.44 Kg

Explanation:

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Energy = mass x latent heat of vaporization

mass = \dfrac{Q}{L_v}

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4 0
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If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon-population sy
Genrish500 [490]

The question is incomplete. The complete question is :

If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon population system move? Assume the population is 7 billion, the average human has a mass of 65 kg, and that the population is evenly distributed over both the Earth and the Moon. The mass of the Earth is 5.97×1024 kg and that of the Moon is 7.34×1022 kg. The radius of the Moon’s orbit is about 3.84×105 m.

Solution :

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Mass of earth, $M_e = 5.97 \times 10^{24} \ kg$

Mass of moon, $M_m = 7.34 \times 10^{22} \ kg$

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Let the earth is at the origin of the coordinate system. Then,

Since $M_e>> M_p$

         $M_m>> M_p$

Hence if we shift all the population on the moon there will be negligible change in the mass of the moon and earth. Hence there will not be any significant shift on the centre of mass. i.e.

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              $= 4.68 \times 10^6 \ m$

$ 4.68 \times 10^3 \ km$ from the earth.

         

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