Question

A box of mass 20 kg is being pulled across the floor with a constant velocity using a force
of 100 N acting at 30o above the horizontal. Calculate the coefficient of kinetic friction

Answer 1

Answer on Question #40925, Physics, Mechanics | Kinematics | Dynamics

Question:

A box of mass $20\mathrm{kg}$ is being pulled across the floor with a constant velocity using a force of $100\mathrm{N}$ acting at 30o above the horizontal. Calculate the coefficient of kinetic friction

Answer:

$F_{fr}$ - friction force

$F$ - pulling force

Newton's first law of motion on y-axis:

$$F \sin 3 0 {}^ {\circ} + N = m g$$ $$F _ {f r} = F \cos 3 0 {}^ {\circ}$$

Therefore, normal force equals:

$$N = m g - F \sin 3 0 {}^ {\circ}$$

Force of friction equals:

$$F _ {f r} = \mu N = \mu (m g - F \sin 3 0 {}^ {\circ})$$

The coefficient of kinetic friction equals:

$$\mu = \frac {F _ {f r}}{(m g - F \sin (3 0 {}^ {\circ}))} = \frac {F \cos 3 0 {}^ {\circ}}{m g - F \sin 3 0 {}^ {\circ}} \cong 0. 5 9$$

Answer: 0.59

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