A box of mass 12.6 kg is pulled to the right with a horizontal force of 14 N. If the box moves to the right with a constant velocity of 1.68 m/s, what is the frictional force acting on the box? What is the coefficient of friction for this situation?
Since there is no accleration, 14 N is the friction force. The speed does not matter at all. The coefficioent of friction is 14 N divided by the weight, M g.
The driver of a 1024 kg truck traveling 32.6 m/s North on I-77 has to make a sudden stop. If it requires a distance of 57.1 m for the truck to stop, use conservation of energy to find the net force acting on the truck.
Cory's question was posted and answered later, in a separate thread
To find the frictional force acting on the box, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
In this case, the box is moving with a constant velocity, so its acceleration is zero. Therefore, the net force acting on the box is also zero. We can write this as:
Net force = 0
The net force acting on the box is the sum of the applied force (14 N) and the frictional force (which is what we are trying to find). So we can write the equation as:
14 N - Frictional force = 0
Rearranging the equation, we can isolate the frictional force:
Frictional force = 14 N
So the frictional force acting on the box is 14 N.
To find the coefficient of friction, we can use the formula:
Coefficient of friction = Frictional force / Normal force
The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, since the box is on a horizontal surface, the normal force is equal to the weight of the box, which can be calculated as:
Weight = mass * gravity
Where gravity is the acceleration due to gravity, approximately 9.8 m/s².
Weight = 12.6 kg * 9.8 m/s² = 123.48 N
Now we can substitute the values into the formula for the coefficient of friction:
Coefficient of friction = 14 N / 123.48 N
Calculating this, we get:
Coefficient of friction ≈ 0.113
Therefore, the coefficient of friction for this situation is approximately 0.113.