Answer:
The correct response is "54.19 kg".
Explanation:
The given values are:
Force,
F = 500 N
Coefficient of static friction,
μ = 0.61
Angle,
[tex]\Theta = 25^{\circ}[/tex]
As we know,
⇒ [tex]F_x=F= \mu_s.N[/tex]
then,
⇒ [tex]FCos \Theta=\mu_s(mg+FSin \Theta)[/tex]
On substituting the given values, we get
⇒ [tex]500Cos(25^{\circ})=0.61[m(9.81)+500 Sin25^{\circ}][/tex]
On substituting the values of Cos and Sin, we get
⇒ [tex]453.154=0.61(9.81m+211.31)[/tex]
⇒ [tex]453.154=5.9842m+128.8991[/tex]
⇒ [tex]453.154-128.8991=5.9842m[/tex]
⇒ [tex]324.2549=5.9842m[/tex]
⇒ [tex]m=\frac{324.2549}{5.9842}[/tex]
⇒ [tex]m=54.19 \ kg[/tex]
