Respuesta :
Answer:
Explanation:
Given
mass of box [tex]m=10\ kg[/tex]
Force applied [tex]F=40 \N[/tex]
inclination of force [tex]\theta =30^{\circ}[/tex] (w.r.t horizontal)
coefficient of kinetic friction [tex]\mu _k=0.3[/tex]
Net Normal reaction on the box is
[tex]N=mg-F\sin \theta [/tex]
Friction force [tex]f_r=\mu _kN[/tex]
[tex]f_r=0.3(10\times 9.8-40\times \sin 30)[/tex]
[tex]f_r=23.4\ N[/tex]
force which is moving the box
[tex]F\cos \theta [/tex]
Net force on the block
[tex]F\cos \theta -f_r=ma[/tex]
[tex]40\times \cos 30-23.4=10\times a[/tex]
[tex]a=1.12\ m/s^2[/tex]
Answer:
1.66 m/s²
Explanation:
mass of box, m = 10 kg
Force, F = 40 N
Angle of inclination, θ = 30°
coefficient of friction, μ = 0.3
Let N be the normal reaction,
equilibrium of force in y axis
N + F Sinθ = mg
N = 10 x 9.8 - 40 Sin 30
N = 98 - 20 = 78 N
Friction force, f = μN = 0.3 x 78 = 23.4 N
Now, use Newton's second law in X axis
F - f = ma
where, a be the acceleration
40 - 23.4 = 10 x a
a = 1.66 m/s²
Thus, the acceleration is 1.66 m/s².
