Question

A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a constant force of 2500 N and the coefficient of kinetic friction between the ground and the log is 0.45, determine the time for the log to reach a speed of 0.5 m/s.

Answer 1

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

$F_f = \mu mg$

where

$\mu=0.45$ is the coefficient of friction

m = 300 kg is the mass of the log

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$F_f = (0.45)(300)(9.8)=1323 N$

So the net force acting on the log is

$F=2500 - 1323=1177 N$

Now, we can find the acceleration of the log by using Newton's second law

$F=ma$

where a is the acceleration. Re-arranging for a,

$a=\frac{F}{m}=\frac{1177}{300}=3.92 m/s^2$

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:

$v=u+at$

where u = 0 is the initial speed and t the time. Solving for t,

$t=\frac{v}{a}=\frac{0.5}{3.92}=0.128 s$