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 2850 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

0.1 s

Explanation:

The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log

So F - f = ma

F - μmg = ma

F/m - μg = a

So, substituting the values of the variables into the equation, we have

a = F/m - μg

a = 2850 N/300 kg - 0.45 × 9.8 m/s²

a = 9.5 m/s² - 4.41 m/s²

a = 5.09 m/s²

Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.

So, making t subject of the formula, we have

t = (v - u)/a

substituting the values of the variables into the equation, we have

t = (v - u)/a

t = (0.5 m/s - 0 m/s)/5.09 m/s²

t = 0.5 m/s ÷ 5.09 m/s²

t = 0.098 s

t ≅ 0.1 s

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