Question

A 5.0 kg block is pushed 2.0 m at a con-

Answer 1

Answer:

$W_F=127.64283 J$

Explanation:

Information Given:

$m = 5kg$  $v=constant$

Key: μ = Kinetic Friction (Kf)   θ = Theta α = 180° N = Normal Force

$W_F=F_ydcos$θ

$W_F=Fdsin$θ

$_{net}F_y = sin$θ-μ$N-mg=0$

$_{net}F_x = 0$

$N=Fcos$θ

$Fsin$θ-μ$N=mg$

$Fsin$θ-μ$Fcos$θ$=mg$

$F=\frac{mgdsin(theta)}{sin(theta)-(Kf)cos(theta)}$ →$W_F=\frac{mgdsin(theta)}{sin(theta)-(Kf)cos(theta)}$

$W_F=\frac{(2)(9.81)(2)sin(30)}{sin(30)-(0.40)cos(30)}$

$W_F=127.64283 J$