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Sagot :
Friction force opposes relative motion between two bodies in contact
The coefficient of friction between the box and the floor is 1.456 × 10⁻²
Reason:
Known parameter are;
Mass of the toy box = 7.0 kg
Velocity of the toy box as of is pushed = Constant velocity
Force applied by the toddler, [tex]F_{toddler}[/tex] = 35 N
Required:
To calculate the coefficient of kinetic friction between the box and the floor
Solution:
Given that the box moves with constant velocity, the net force acting on the box is zero, which gives;
[tex]F_{Net} = F_{toddler} - F_{friction} = 0[/tex]
Where;
[tex]F_{friction}[/tex] = Friction force
[tex]\therefore F_{friction} = F_{toddler} = \mathbf{35 \, N}[/tex]
[tex]F_{friction}[/tex] = Normal reaction, N × Coefficient of friction, [tex]\mu_k[/tex] = N·[tex]\mu_k[/tex]
On a flat surface, normal reaction of an object = The weight of the object
∴ [tex]F_{friction}[/tex] = Weight of box × Coefficient of friction
[tex]Coefficient \ of \ friction, \ \mu_k = \dfrac{ F_{friction} }{Weight}[/tex]
Weight of the box, W = 7.0 kg × 9.81 m/s² = 68.67 N
The coefficient of friction that gives a force of 35 N between the box and the floor is therefore;
[tex]Coefficient \ of \ friction, \ \mu_k = \dfrac{ 35 \, N\ }{68.67 \, N} \approx 1.456 \times 10^{-2}[/tex]
The coefficient of friction between the box and the floor, [tex]\mu_k[/tex] = 1.456 × 10⁻²
Learn more about friction force here:
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