Question

A 7.22-N force is applied (Fapp) to a 1.86-kg object to accelerate it rightwards. The object encounters 3.46-N of friction (Ff). Determine the acceleration (a) of the object.

Answer 1

$2.02\text{ }\frac{m}{s^2}$

Explanation

Step 1

Free body diagram

Step 2

Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, a force applied to an object at rest causes it to accelerate in the direction of the force.it is given by the formula

$\begin{gathered} F=ma \\ wheere\text{ F is the resultant of the applied force acting} \\ m\text{ is the mass} \\ a\text{ is the acceleration} \end{gathered}$

so

a)let

$\begin{gathered} m=1.86\text{ Kg} \\ Ff=3.46\text{ N} \\ Applied\text{ force=7.22 N} \end{gathered}$

b) now ,replace and solve for a

from the free body diagram and the data we know that the object accelerates horizontally ,so the sum of the applied force make the object moves rigthwards

$\begin{gathered} F=ma \\ 7.22\text{ N-3.46 N=1.86 kg*a} \\ 3.76\text{ N=1.86 Kg *a} \\ divide\text{ both sides by 1.86 kg} \\ \frac{3.76N}{1.86\text{ Kg}}\text{=}\frac{1.86Kg*a}{1.86\text{ Kg}} \\ 2.02\text{ }\frac{m}{s^2}=a \end{gathered}$

therefore,

the acceleration of the objects is 2.02 meters per square second

$2.02\text{ }\frac{m}{s^2}$

I hope this helps you