Answer:
0.128 s
Explanation:
We have to start by calculating the net force acting on the log. We have two forces:
- The constant pulling force, forward, of F = 2500 N
- The frictional force, backward
The frictional force is given by
[tex]F_f = \mu mg[/tex]
where
[tex]\mu=0.45[/tex] is the coefficient of friction
m = 300 kg is the mass of the log
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Substituting,
[tex]F_f = (0.45)(300)(9.8)=1323 N[/tex]
So the net force acting on the log is
[tex]F=2500 - 1323=1177 N[/tex]
Now, we can find the acceleration of the log by using Newton's second law
[tex]F=ma[/tex]
where a is the acceleration. Re-arranging for a,
[tex]a=\frac{F}{m}=\frac{1177}{300}=3.92 m/s^2[/tex]
And finally we can find the time it takes for the log to reach a speed of
v = 0.5 m/s
by using the suvat equation:
[tex]v=u+at[/tex]
where u = 0 is the initial speed and t the time. Solving for t,
[tex]t=\frac{v}{a}=\frac{0.5}{3.92}=0.128 s[/tex]
