Answer:
W = (F1 - mg sin θ) L, W = -μ mg cos θ L
Explanation:
Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane
Y Axis
N -
=
N = W_{y}
X axis
F1 - fr - Wₓ = 0
fr = F1 - Wₓ
Let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
We substitute
fr = F1 - W sin θ
Work is defined by
W = F .dx
W = F dx cos θ
The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1
W = -fr x
W = (F1 - mg sin θ) L
Another way to calculate is
fr = μ N
fr = μ W cos θ
the work is
W = -μ mg cos θ L
Report this clown who put the first answer he’s trying to get your ip
Explanation:
First, simplify the circuit. Then calculate the parallel and consecutive resistances to find the answer.
Answer: As per the question, a ray of light is incident on a surface and it is partly reflected and refracted. The incident light is an unpolarised light. The reflected light is partially polarised.
If the angle of incidence becomes equal to the Brester angle (polarising angle), then the reflected light is completely plane polarised.