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
\(F=ma\)
\(F=75\times3.60\)
\(F=270\ N\)
Now, we need to calculate the parallel force
\(F'=mg\sin\theta\)
\(F'=75\times9.8\times\sin25^{\circ}\)
\(F'=310.62\ N\)
We calculate the friction force
\(f_{\mu}=F'-F\)
\(f_{\mu}=310.6-270\)
\(f_{mu}=40.6\ N=approx 41\ N\)
If the box is moving in an angle, then the weight
\(W = mg\ \cos\theta\)
If it is horizontally then
\(W = mg\)
So, The normal force is
\(F'' = 75\times9.8\times\cos25^{\circ}\)
\(F''= 666.13\ N\)
We need to calculate the coefficient between the box and ramp
\(f_{\mu}=\mu\times F''\)
Where, \(\mu\)=friction coefficient
\(f_{\mu}\)=frictional force
F''=normal force
Put the value into the formula
\(\mu=\dfrac{f_{\mu}}{F''}\)
\(\mu=\dfrac{41}{666.13}\)
\(\mu=0.061\)
We need to calculate the acceleration
\(F''-f_{\mu}=ma\)
\(mg\cos25^{\circ}-\mu mg=ma\)
\(175\times9.8\cos25.0^{\circ}-0.061\times175\times9.8=175\times a\)
\(a=\dfrac{175\times9.8\cos25.0^{\circ}-0.061\times175\times9.8}{175}\)
\(a=8.284\ m/s^2\)
Hence, The coefficient and acceleration are 0.061 and 8.284 m/s².
