Respuesta :
Answer:
Part a)
[tex]a = 0.36 m/s^2[/tex]
Part b)
[tex]a = -1.29 m/s^2[/tex]
Explanation:
Force applied by the student on the box is 80 N at an angle of 25 degree
so here two components of the force on the box is given as
[tex]F_x = Fcos25[/tex]
[tex]F_x = 80 cos25 = 72.5 N[/tex]
[tex]F_y = Fsin25[/tex]
[tex]F_y = 80 sin25 = 33.8 N[/tex]
now in vertical direction we can use force balance for the box to find the normal force on it
[tex]F_n + F_y = mg[/tex]
[tex]F_n = (25)(9.81) - 33.8[/tex]
[tex]F_n = 211.45 N[/tex]
now kinetic friction on the box opposite to applied force due to rough floor is given as
[tex]F_k = \mu F_n[/tex]
[tex]F_k = (0.300)(211.45) = 63.44 N[/tex]
now the net force on the box in forward direction is given as
[tex]F_{net} = F_x - F_k [/tex]
[tex]F_{net} = 72.5 - 63.44 = 9.065 N[/tex]
now the acceleration of the box is given as
[tex]a = \frac{F_{net}}{m}[/tex]
[tex]a = \frac{9.065}{25} = 0.36 m/s^2[/tex]
Part b)
when box is pulled up along the inclined surface of angle 10 degree
now the two components of the force will be same along the inclined and perpendicular to inclined plane
[tex]F_x = 72.5 N[/tex]
[tex]F_y = 33.8 N[/tex]
now force balance perpendicular to inclined plane is given as
[tex]F_n + F_y = mgcos\theta[/tex]
[tex]F_n = (25)(9.81)cos10 - 33.8 = 207.7 N[/tex]
now the friction force opposite to the motion on the box is given as
[tex]f_k = \mu F_n[/tex]
[tex]f_k = (0.300)(207.7) = 62.3 N[/tex]
now the net pulling force along the inclined plane is given as
[tex]F_{net} = F_x - F_k - mgsin10[/tex]
[tex]F_{net} = 72.5 - 62.3 - (25)(9.81)sin10[/tex]
[tex]F_{net} = -32.38 N[/tex]
now the box will decelerate and it is given as
[tex]a = \frac{F_{net}}{m}[/tex]
[tex]a = \frac{-32.38}{25} = -1.29 m/s^2[/tex]
The acceleration of the box when the student was it along the horizontal is 0.37 m/s².
The acceleration of the box when it was pulled up 10Ⱐincline is -1.29 m/s².
The given parameters;
- mass of the box, m = 25 kg
- force applied on the box, F = 80 N
- angle of inclination of the rope, θ = 25 â°
The normal force on the box is calculated as;
[tex]F_n = W - Fsin(\theta)\\\\F_n = (25\times 9.8) - (80 sin(25))\\\\F_n = 211.19 \ N[/tex]
The frictional force on the box is calculated as;
[tex]F_k = \mu_k F_n\\\\F_k = 0.3\times 211.19 \\\\F_k = 63.36 \ N[/tex]
The acceleration of the box is calculated as;
[tex]Fcos(\theta) - F_k = ma\\\\a = \frac{Fcos(\theta) - F_k}{m} \\\\a = \frac{80\times cos(25) - 63.36}{25} \\\\a = 0.37 \ m/s^2[/tex]
When the student starts pulling the box at 10.0° incline;
The normal force on the box is calculated as;
[tex]F_n = Wcos(\theta) - Fsin(\theta)\\\\F_n = (25\times 9.8 \times cos(10)) - (80 sin(25))\\\\F_n = 207.27 \ N[/tex]
The frictional force on the box is;
[tex]F_k = \mu_k F_n\\\\F_k = 0.3\times 207.27 \\\\F_k = 62.18 \ N[/tex]
The acceleration of the box is calculated as;
[tex]Fcos(25) - F_k - Wsin(10) = ma\\\\(80cos(25) ) - 62.18 - (25\times 9.8\times sin(10)) = 25a\\\\-32.23 =25a\\\\a = \frac{-32.23}{25} \\\\a = -1.29 \ m/s^2[/tex]
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