Question

Una caja de 40.0 kg se desliza 4.00 m hacia arriba por un plano inclinado de 26° con respecto a la acción de una fuerza de 340 N paralela al plano. Si el coeficiente de fricción cinético entre las superficies es 0.250, calcula el trabajo neto realizado sobre el objeto.

Answer 1

Answer:

W = 320.30 J

Explanation:

To calculate the net work done over the block you take into account all implied forces:

$\Sigma F=F+F_gsin\theta-F_fcos\theta$  (1)

The gravitational force and friction force are against the applied force F.

θ = 26°

F: applied force = 340N

Fg: gravitational force = Mg = (40.0kg)(9.8m/s^2) = 392N

Ff: friction force = $\mu N=\mu Mg=(0.250)(392N)=98N$

Next, you replace to obtain the net force:

$F_N=(340N)-(392N)sin(26\°)-(98N)cos(26\°)\\\\F_N=80.07N$

Finally, the net work, for 4 m, is:

$W_N=F_Nd=(80.07N)(4m)=320.30J$