# please help with question below?

You need to move a 680 kg log with your tractor. The coefficient of static friction between the log and the ground is µs=0.700, and the coefficient of kinetic friction is µk=0.600. Your tractor applies a horizontal force of magnitude F=5.0 kN to the log. What is the acceleration of the log? You can assume that the gravitational acceleration g=10m/s2.

### 2 Answers

- oubaasLv 73 weeks ago
once it starts moving :

a = (F-m*g*µs)/m = (5000-680*10*0.7)/680 = 0.35 m/sec^2 (just 2 sign. digits)

once in motion :

a' = (F-m*g*µk)/m = (5000-680*10*0.6)/680 = 1.4 m/sec^2 (just 2 sign. digits)

- NCSLv 73 weeks ago
First, you must determine if the static friction force is overcome.

max fs = µs*m*g = 0.700 * 680kg * 10m/s² = 4760 N

which is < 5000 N applied, so yes.

a = net F / mass = (5000 - 0.600*680kg*10m/s²) / 680kg

a = 1.35 m/s²

If you find this helpful, please select Favorite Answer!