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:

Show me the final answer↓

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 N

 

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

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

F_{net}=170.63 N

 

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.

Related Questions

Leave a comment

Your email address will not be published. Required fields are marked *