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marta [7]
3 years ago
14

When a 9.9 kg object is in free fall, it feels a force of 97.02 N. what is its acceleration?

Physics
1 answer:
Free_Kalibri [48]3 years ago
6 0

The acceleration of an object in free fall is 9.8 m/s^2

Explanation:

We can solve the problem by using Newton's second law of motion, which states that:

F=ma

where

F is the net force on an object

m is the mass of the object

a is its acceleration

For the object in this problem, we have:

m = 9.9 kg is its mass

F = 97.02 N is the net force on the object

So, we can rearrange the equation to find its acceleration:

a=\frac{F}{m}=\frac{97.02}{9.9}=9.8 m/s^2

This value is actually known as acceleration of gravity (g), and it is the acceleration that any object in free fall has near the Earth's surface, regardless of its mass.

Learn more about Newton's second law here:

brainly.com/question/3820012

And about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

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The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

    F_{1} = 20 N, F_{2} = 25 N, a = -0.9 m/s^{2}

             W = 83 N

         m = \frac{83}{9.81}

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Now, we will balance the forces along the y-component as follows.

       N = W + F_{2}

           = 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

       F_{1} - F_{r} = ma

        20 - F_{r} = 8.46 \times (-0.9)

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Also, we know that relation between force and coefficient of friction is as follows.

             F_{r} = \mu \times N

          \mu = \frac{F_{r}}{N}

                    = \frac{7.614}{108}

                    = 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

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