If the sled in the the previous question weighs 530 newtons and the horse pulls it with a force of 140 newtons across level ground at a constant speed, what is the magnitude of the force of friction between the sled rails and the snow?
Hide hint for Question 8
Hint 1: Remember that "level ground at constant speed" are magic words, what do they tell you about the forces?
Hint 2: Be sure you understand the different concepts, "force of friction" and "coefficient of friction". You have enough information to give either of these, but this question merely asks for the force.
constant speed, no acceleration.
pulling force-friction=mass*acceleration
To find the magnitude of the force of friction between the sled rails and the snow, we need to use Newton's second law of motion, which states that the sum of all the forces acting on an object is equal to the mass of the object multiplied by its acceleration.
In this case, the sled is moving at a constant speed, which means there is no acceleration. Therefore, the net force acting on the sled must be zero.
The net force acting on the sled can be calculated by subtracting the force of friction from the force applied by the horse. So, we can use the equation:
Force applied by the horse - Force of friction = Net force
We are given that the force applied by the horse is 140 newtons, and we need to find the force of friction. The net force is zero since the sled is moving at a constant speed.
Rearranging the equation, we have:
Force of friction = Force applied by the horse - Net force
Since the net force is zero, we can simplify the equation to:
Force of friction = 140 newtons - 0 newtons
Therefore, the magnitude of the force of friction between the sled rails and the snow is 140 newtons.