A diagram represents a 5 N force to the North and a 12 N force East acting on an object P of mass 6.5kg.

-The Magnitude of the Net Force is 13 N
- The direction it accelerates is 23 degrees North of East.

FIND:
(A) If the object has an intial speed of 8 m/s in the direction opposite of the net force. How long does it take for the object to come to a stop?

________

F=ma

a=F/m = 13/6.5 =2 m/s²
v=v₀-at
v=0
v₀-at=0 =>t=v₀/a=8/2 = 4 s

To find the time it takes for the object to come to a stop, we can use the equation of motion:

\( v = u + at \),

where
- \( v \) is the final velocity (which is 0 m/s since the object comes to a stop),
- \( u \) is the initial velocity (-8 m/s as it is in the opposite direction of the net force),
- \( a \) is the acceleration of the object,
- and \( t \) is the time.

We can find the acceleration of the object using the net force, mass, and the equation \( F = ma \):

\( F = ma \)
\( 13\,N = 6.5\,kg \cdot a \)
\( a = \frac{13\,N}{6.5\,kg} \)
\( a = 2\,m/s^2 \)

Now, substituting the values into the equation of motion:

\( 0\,m/s = -8\,m/s + 2\,m/s^2 \cdot t \)

To solve for \( t \), we can rearrange the equation as follows:

\( 8\,m/s = 2\,m/s^2 \cdot t \)
\( t = \frac{8\,m/s}{2\,m/s^2} \)
\( t = 4\,s \)

Therefore, it will take 4 seconds for the object to come to a stop.