A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exerted downward at an angle of 35.2 degrees below the horizontal. Fink Mu,k between the box and the floor.
There have been several of this type of question posted here today. Take a look at how they were done.
Each can be solved by writing down Newton's second law. The force "F" in the law F= m a is the net force. In this case, the acceleration a is zero. Therefore the applied force component along the direction of motion equals the friction force. Write that statement in algebraic form and solve for Mu,k.
1. The 425 N force needs to be broken into component vectors. 425*sin35.2 will give you the vertical component, 425*cos35.2 will give you the horizontal component.
2. The vertical component gets added to the box's 325 N weight because it is being pushed down on.
3. The horizontal component is equaled by the force of friction backward. Divide this force by the force from step 2 to get the coefficient of friction, which is about 0.61.
To find the coefficient of kinetic friction (μk) between the box and the floor, we need to use Newton's second law and the equation for friction.
1. Newton's second law states that the net force (Fnet) acting on an object is equal to the mass of the object (m) multiplied by its acceleration (a). In this case, the box is moving at a constant velocity, so its acceleration is zero. Therefore, the net force on the box is also zero.
2. The net force can be calculated by summing up all the forces acting on the box. In this problem, the only horizontal force acting on the box is the friction force (Ff).
3. The equation for friction force (Ff) is given by Ff = μk * Fn, where μk is the coefficient of kinetic friction and Fn is the normal force.
4. In this problem, the applied force (425 N) is exerted downward at an angle of 35.2 degrees below the horizontal. We can separate this force into its horizontal and vertical components.
The horizontal component of the applied force is Fh = F * cos(35.2°), where F is the magnitude of the applied force.
Fh = 425 N * cos(35.2°)
The vertical component of the applied force is Fv = F * sin(35.2°), where F is the magnitude of the applied force.
Fv = 425 N * sin(35.2°)
5. Since the box is in equilibrium and not accelerating vertically, the vertical component of the applied force (Fv) must be balanced by the normal force (Fn) exerted by the floor upward. Therefore, Fn = Fv = 425 N * sin(35.2°).
6. Now, we can substitute the value of Fn into the equation for friction force: Ff = μk * Fn.
Ff = μk * (425 N * sin(35.2°))
7. Since the box is moving at a constant velocity, the friction force (Ff) must be equal in magnitude to the applied horizontal force (Fh). Therefore, Ff = Fh.
8. Now, we can set up the equation Ff = Fh and solve for the coefficient of kinetic friction (μk):
μk * (425 N * sin(35.2°)) = 425 N * cos(35.2°)
Simplifying the equation, μk * sin(35.2°) = cos(35.2°)
Dividing both sides of the equation by sin(35.2°), we get:
μk = cos(35.2°) / sin(35.2°)
9. Evaluating this expression, you can use a calculator to find the value of μk.
By following these steps, you can find the coefficient of kinetic friction (μk) between the box and the floor.