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2 years ago

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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?

1 answer:

olya-2409 [2.1K]2 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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