A 3-kg rock is thrown upward with a force

A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 $m/s^2$ . Determine the acceleration of the rock, in $m/s^2$ .

Solution:

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Let us first calculate the weight of the rock using the following equation:

W=mg

(Where $W$ is weight, $m$ is mass and $g$ is acceleration due to gravity)

Substitute the values we know:

W=mg

W=(3)(9.79)

W=29.37

We will now calculate all the forces that affect the rock (net force):

\uparrow^+ F_{net}=200-29.37

F_{net}=170.63

Let us now calculate the acceleration using Newtons second law.

F=ma

(Where $F$ is force, $m$ is mass and $a$ is acceleration)

Substituting the values we know gives us:

F=ma

(isolate for $a$ )

$a=\dfrac{F}{m}$

$a=\dfrac{170.63}{3}$

$a=56.88\,\text{m}/\text{s}^2$

Final Answer:

a=56.88\,\text{m}/\text{s}^2

This question can be found in Thermodynamics, an Engineering Approach, 8th edition, Chapter 1, question 1-12.

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Solution:

Final Answer:

This question can be found in Thermodynamics, an Engineering Approach, 8th edition, Chapter 1, question 1-12.

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