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AnnZ [28]
3 years ago
10

Two identical objects, A and B, are sitting on a table. If the net force on object A is 5 N and the net force on object B is 10

N, which of the following statements is true?
A. Neither object will accelerate.
B. Object A will accelerate more quickly than object B.
C. Object B will accelerate more quickly than object A.
D. The acceleration of each object will depend on other forces than the net force.
Physics
1 answer:
sveta [45]3 years ago
7 0

If the net force on object A is 5 N and the net force on object B is 10 N, then object B will accelerate more quickly than object A provided the mass of both objects are same.

Answer: Option C

<u>Explanation: </u>

According to Newton’s second law of motion, any external force applied on an object is directly proportional to the mass and acceleration of the object. In order to state this law in terms of acceleration, it is stated that acceleration exhibited by any object is directly proportional to the net force applied on the object and inversely proportional to the mass of the object as shown below:

                      \text {Acceleration of the object } \propto \frac{\text {Net force on the object}}{\text {Mass of the object}}

So if two objects A and B are identical which means they have same mass, then the acceleration attained by the object will be directly proportionate to the net forces exerted on the objects only.

Thus if the force applied is more for one object, then the object will be exhibiting more acceleration compared to the other one. So as object B is experiencing a net force of 10 N which is greater than the net force experiences by object A, then the object B will be accelerating more quickly compared to the object A's acceleration.

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2 years ago
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
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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

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3 years ago
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8. Il An 8.00 kg package in a mail-sorting room slides 2.00 m down a
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Answer:

See below

Explanation:

Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N

  no work is done by this force

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   work of friction = 7.55 * 2 m = 15.1 j

Force Downplane = mg sin 53 = 62.61  N

    work = 62.61 * 2 = 125.22 j

Net Force downplane =   force downplane - force friction = 55.06 N

net Work = force * distance = 55.06 N * 2 M = 110.12 j

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