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Ksivusya [100]
3 years ago
7

A 5.00-kg object is initially at rest. The object is acted on by a 9.00-N force toward the east for 3.00 s. No force acts on the

object for the next 4.00 s. How far has the object moved during this 7.00 s interval?
Physics
1 answer:
olya-2409 [2.1K]3 years ago
7 0

Answer:

The total distance at 7 s is:

x_{tot}=27\: m

Explanation:

<u>Distance due to the force</u>

We can use second Newton's law to find the acceleration.

F=ma

a=\frac{F}{m}=\frac{9}{5}=1.8\: m/s^{2}

Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.

x_{1}=0.5at_{1}^{2}

x_{1}=0.5(1.8)(3)^{2}

x_{1}=8.1\: m

In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.

First, we need to find the final velocity of the first interval

v=v_{i}+at_{1}=0+(1.8)3=5.4\: m/s

So the second distance will be:

x_{2}=vt_{2}=5.4*4=21.6\: m

Therefore, the total distance is:

x_{tot}=x_{1}+x_{2}=5.4+21.6=27\: m

I hope it helps you!

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