What average acceleration es the sled experience?

[adv]

1. Six dogs, each having a mass of 32 kg, pull a 325 kg sled horizontally across ice. If each dog applies a force of 86 N [fwd], what average acceleration es the sled experience? Assume that friction is negligible.


Equation:
F=ma

My solution:

6 x 86 N= 516 N

F=ma
a= F/m
a= 516 N [fwd]/ 325 kg
a= 1.588 m/s2
a= 1.6 m/s2

I just have a problem understanding the question. Help please!

 

[Q_Goest]

Your approach is good, but you didn't account for the mass of the dogs. Don't forget that the total mass of the dogs is also being accelerated.

 

[adv]

The force of the dogs, 6 x 86 N= 516 N divided by the mass of the sled plus the mass of the dogs gives the average acceleration of the sled? So the acceleration of the sled equals the acceleration of the dogs.

F=ma
a= F/m
a= 516 N [fwd]/ 517 kg
a= 0.998 m/s2
a= 1.0 m/s2