Respuesta :
Answer:
The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.
Explanation:
Given that,
Mass = 12.0 kg
Angle = 31.0°
Friction coefficient = 0.158
Mass of second block = 38.0 kg
Using formula of frictional force
[tex]f_{\mu} = \mu N[/tex]....(I)
Where, N = normal force
[tex]N = mg\cos\theta[/tex]
Put the value of N into the formula
[tex]N =12\times9.8\times\cos 31^{\circ}[/tex]
[tex]N=100.80\ N[/tex]
Put the value of N in equation (I)
[tex]f_{mu}=0.158\times100.80[/tex]
[tex]f_{mu}=15.9264\ N[/tex]
Now, Weight of second block
[tex]W = mg[/tex]
[tex]W=38.0\times9.8[/tex]
[tex]W=372.4\ N[/tex]
The horizontal force is
[tex]F = mg\sintheta[/tex]
[tex]F=12\times9.8\times\sin 31^{\circ}[/tex]
[tex]F=60.5684\ N[/tex]....(II)
(I). We need to calculate the acceleration
[tex]a=m_{2}g-\dfrac{f_{\mu}+mg\sin\theta}{m_{1}+m_{2}}[/tex]
[tex]a=\dfrac{372.4-(15.9264+60.5684)}{12+38}[/tex]
[tex]a=5.92\ m/s^2[/tex]
(II). We need to calculate the tension in the rope
[tex]m_{2}g-T=m_{2}a[/tex]
[tex]-T=38\times5.92-38\times9.8[/tex]
[tex]T=147.44\ N[/tex]
Hence, The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.
