Respuesta :
Answer:
0.6
Explanation:
There are four forces on the box:
Weight force mg pulling down,
Normal force N pushing up,
Tension force T pulling 30° above the horizontal,
and friction force Nμ pushing to the left.
Sum of the forces in the y direction:
∑F = ma
N + T sin 30° − mg = 0
N = mg − T sin 30°
Sum of forces in the x direction:
∑F = ma
T cos 30° − Nμ = 0
Nμ = T cos 30°
μ = T cos 30° / N
μ = T cos 30° / (mg − T sin 30°)
Plug in values:
μ = (250 N) cos 30° / ((50 kg) (9.8 m/s²) − (250 N) sin 30°)
μ = 0.59
Rounded to one significant figure, the coefficient of friction is 0.6.
The coefficient of kinetic friction between the box and floor is : 0.6
Given data :
Mass of box = 50 kg
Angle = 30°
Force required on rope ( T ) = 250 N
speed = 20 m/s
g = 9.81 m/s²
Calculate the value of the coefficient of kinetic friction
First step : determine sum of forces in the y-direction
∑ Fy = ma
N + T sin 30° - mg = 0
∴ N = mg - Tsin 30° ----- ( 1 )
Next step : Determine the sum of forces in the x-direction
∑ Fx = ma
T cos 30° - Nμ = 0
∴ μ = T cos 30° / N ----- ( 2 )
where ; N = mg - T sin 30°
Final step : Determine the coefficient of kinetic friction between the box and floor
Back to equation ( 2 )
μ = T cos 30° / ( mg - T sin 30° )
= 250 cos 30° / ( ( 50 * 9.81 ) - 250 * sin 30° ))
= 0.59 ≈ 0.6
Hence we can conclude that The coefficient of kinetic friction between the box and floor is : 0.6
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