Respuesta :
Answer:
The coefficient and acceleration are 0.061 and 8.284 m/s².
Explanation:
Given that,
Mass = 75 kg
Angle = 25.0°
Acceleration = 3.60 m/s²
We need to find the force
[tex]F=ma[/tex]
[tex]F=75\times3.60[/tex]
[tex]F=270\ N[/tex]
Now, we need to calculate the parallel force
[tex]F'=mg\sin\theta[/tex]
[tex]F'=75\times9.8\times\sin25^{\circ}[/tex]
[tex]F'=310.62\ N[/tex]
We calculate the friction force
[tex]f_{\mu}=F'-F[/tex]
[tex]f_{\mu}=310.6-270[/tex]
[tex]f_{mu}=40.6\ N=approx 41\ N[/tex]
If the box is moving in an angle, then the weight
[tex]W = mg\ \cos\theta[/tex]
If it is horizontally then
[tex]W = mg[/tex]
So, The normal force is
[tex]F'' = 75\times9.8\times\cos25^{\circ}[/tex]
[tex]F''= 666.13\ N[/tex]
We need to calculate the coefficient between the box and ramp
[tex]f_{\mu}=\mu\times F''[/tex]
Where, [tex]\mu[/tex]=friction coefficient
[tex]f_{\mu}[/tex]=frictional force
F''=normal force
Put the value into the formula
[tex]\mu=\dfrac{f_{\mu}}{F''}[/tex]
[tex]\mu=\dfrac{41}{666.13}[/tex]
[tex]\mu=0.061[/tex]
We need to calculate the acceleration
[tex]F''-f_{\mu}=ma[/tex]
[tex]mg\cos25^{\circ}-\mu mg=ma[/tex]
[tex]175\times9.8\cos25.0^{\circ}-0.061\times175\times9.8=175\times a[/tex]
[tex]a=\dfrac{175\times9.8\cos25.0^{\circ}-0.061\times175\times9.8}{175}[/tex]
[tex]a=8.284\ m/s^2[/tex]
Hence, The coefficient and acceleration are 0.061 and 8.284 m/s².
The coefficient of friction is 0.14 and the acceleration of the 175 Kg box is 2.95 m/s^2.
We know that;
ma = -Ff + mgsinθ
Ff = frictional force on the box
But Ff = μmgcosθ
Hence;
ma = mgsinθ - μmgcosθ
μmgcosθ = mgsinθ - ma
μ = mgsinθ - ma/mgcosθ
Substituting values;
m = 75 kg
a = 3.60 m/s^2
θ = 25.0°
μ =[tex][75 kg * 10 m/s^2 (sin 25.0)] - [75 kg * 3.60 m/s^2]/[75 kg * 10 m/s^2 (cos 25.0)][/tex]
μ = 316.96 N - 225 N/679.73
μ = 91.96/679.73
μ = 0.14
From μmgcosθ = mgsinθ - ma
a = mgsinθ - μmgcosθ /m
a =[tex][175 kg * 10 m/s^2 (sin 25.0)] - [0.14 * 175 kg * 10 m/s^2* (cos 25.0)]/175Kg[/tex]
a = 739.58 - 222.05/175
a = 2.95 m/s^2
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