Answer:
[tex]-1.398\ \text{m/s}^2[/tex]
Decelerating or slowing down
Explanation:
F = Force = 50 N
[tex]\theta[/tex] = Angle force is being applied = [tex]42^{\circ}[/tex]
[tex]\mu[/tex] = Coefficient of friction = 0.64
m = Mass of wagon = 12 kg
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Normal force is given by
[tex]N=mg-F\sin\theta[/tex]
Frictional force is given by
[tex]f=\mu N\\\Rightarrow f=\mu (mg-F\sin\theta)[/tex]
The force balance is given by
[tex]F\cos\theta-f=ma\\\Rightarrow \dfrac{F\cos\theta-\mu (mg-F\sin\theta)}{m}=a\\\Rightarrow a=\dfrac{50\times \cos42^{\circ}-0.64(12\times 9.81-50\times\sin42^{\circ})}{12}\\\Rightarrow a=-1.398\ \text{m/s}^2[/tex]
The acceleration of the wagon is [tex]-1.398\ \text{m/s}^2[/tex]. The negative sign indicates that the wagon is decelerating or slowing down.
The acceleration of the wagon is [tex]-1.398 \;\rm m/s^{2}[/tex].
Given data:
The magnitude of pulling force is, F = 50 N.
The angle of inclination is, [tex]\theta = 42^{\circ}[/tex].
The mass of wagon wheel is, m = 12 kg.
Coefficient of friction between wagon and grass is, [tex]\mu =0.64[/tex].
The given problem is based on the concept of frictional force. The standard expression for the frictional force is,
[tex]f= \mu \times N[/tex]
Here, N is the normal force and its value is,
[tex]N=mg-Fsin \theta[/tex]
And the net force acting on wagon is,
[tex]F' = Fcos\theta -f\\\\ma = Fcos\theta -(\mu(mg-Fsin \theta))\\\\a = \dfrac{Fcos\theta -(\mu(mg-Fsin \theta))}{m}[/tex]
Here, a is the acceleration of wagon.
Solving as,
[tex]a = \dfrac{50 \times cos42 -(0.64(12 \times 9.8-(50 \times sin42)))}{12}\\\\a=-1.398 \;\rm m/s^{2}[/tex]
Thus, we can conclude that the acceleration of the wagon is [tex]-1.398 \;\rm m/s^{2}[/tex].
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