An 103 Newtons box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at an acceleration of -0.9m/s2. The push force has a horizontal component of 20N and a vertical component of 25N downward. Calculate the coefficient of the kinetic friction and the box. Hint. Draw all the forces. Apply Newtons 2nd law : ƩFy to solve for the normal force and ƩFx to solve for the frictional force. Then apply the formula : frictional force = µkN.
M*g = 103 N. = Wt. of box.
M = 103/g = 103/9.8 = 10.5 kg.
Fn = 103*Cos 0 + 25 = 128 N. = normal force.
Fk = u*Fn = 128u.
Fx-Fk = M*a.
20-128u = 10.5*(-0.9), u = ?.
To solve this problem, we need to apply Newton's second law and consider the various forces acting on the box.
1. Draw the forces:
First, let's draw the forces acting on the box:
- The weight (mg) acting vertically downward, where m is the mass of the box and g is the acceleration due to gravity.
- The normal force (N) acting perpendicular to the floor.
- The push force with its horizontal component (20N) and vertical component (25N downward).
- The frictional force (f) opposing the motion of the box.
Now, let's break down these forces into their x and y components:
- The weight can be broken down into its vertical component (mg) and no horizontal component since it is acting vertically.
- The normal force has no x or y components since it acts perpendicular to the floor.
- The push force has a horizontal component (20N) and a vertical component (25N downward).
2. Apply Newton's second law:
The sum of forces in the x-direction (ƩFx) is equal to the mass of the box multiplied by its acceleration. In this case, the acceleration is given as -0.9 m/s^2 (negative because it is slowing down):
20N - frictional force = m * (-0.9 m/s^2)
The sum of forces in the y-direction (ƩFy) is equal to zero since there is no vertical acceleration:
N - 25N - mg = 0
3. Solve for the unknown variables:
Let's solve these equations to find the normal force (N) and the coefficient of kinetic friction (µk).
From the equation in the y-direction:
N - 25N - mg = 0
N = 25N + mg
Substitute this value of N in the x-direction equation:
20N - frictional force = m * (-0.9 m/s^2)
20(25N + mg) - frictional force = m * (-0.9 m/s^2)
Now, we can solve for the frictional force:
frictional force = 20(25N + mg) + 0.9m
Then, we can use the formula frictional force = µkN to find the coefficient of kinetic friction (µk).
4. Calculate the coefficient of kinetic friction (µk):
Substitute the value of frictional force back into the frictional force equation:
frictional force = µkN
20(25N + mg) + 0.9m = µk(25N + mg)
Simplify the equation by dividing both sides by (25N + mg):
20 + 0.9m/(25N + mg) = µk
Finally, we have the equation to calculate the coefficient of kinetic friction (µk) for the box of oranges.
5. Plug in the known values:
To find the numerical value of µk, you need to know the mass of the box (m) and the value of g.