Question

A 5.0 kg block is pushed 2.0 m at a con-
stant velocity up a vertical wall by a constant
force applied at an angle of 30.0° with the
horizontal, as shown in the figure.
The acceleration of gravity is 9.81 m/s2.
F
30°
2 m
5 kg
Drawing not to scale.
If the coefficient of kinetic friction between
the block and the wall is 0.40, find
a) the work done by the force on the block.
Answer in units of J.

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$=mg$

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

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

$W_F=127.64283 J$