Respuesta :
Answer:
W = (F1 - mg sin θ) L,  W = -Ο  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 Â
    N - [tex]W_{y}[/tex] =
    N = W_{y}
X axis
    F1 - fr - Wâ = 0
    fr = F1 - Wâ
Let's use trigonometry to find the components of the weight
   sin θ = Wâ / W
   cos θ = W_{y} / W
   Wâ = 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
    Â
    W = -fr x
    W = (F1 - mg sin θ) L
Another way to calculate is
     fr = Ο N
     fr = Ο W cos θ
the work is
     W = -Ο  mg cos θ L
The normal reaction prevents the block from sinking into the inclined plane
Work done by friction on the block is  Ο¡L¡cos(θ)(Fâ¡sin(θ) - m¡g)
The reason above expression for work done is correct is presented as follows:
The given parameters are;
The magnitude of the force acting on the block = Fâ
Coefficient of kinetic friction between the plane and the block = Îź
Work done by the force of friction = [tex]W_{fric}[/tex]
The angle of inclination of the plane = θ
Distance up the incline the block moves = L
Required:
The work done on the block by the force of friction, [tex]W_{fric}[/tex]
Solution:
Normal reaction, N = (m¡g - Fâ¡sin(θ))¡cos(θ)
Fictional force, [tex]F_f[/tex] = -Ο¡(m¡g - Fâ¡sin(θ))¡cos(θ) (opposite in direction to the applied force)
Work done = Force Ă Direction of the force
Work done on the block by friction, as the block moves up the inclined plane a distance, L is [tex]W_{fric}[/tex] = [tex]F_f[/tex] Ă L
â´ [tex]W_{fric}[/tex] =  -Ο¡(m¡g - Fâ¡sin(θ))¡cos(θ) Ă L
Work done on the block by friction, as the block moves up the inclined plane a distance, L, [tex]W_{fric}[/tex] =  -Ο¡(m¡g - Fâ¡sin(θ))¡cos(θ) Ă L= Ο¡L¡Fâ¡cos(θ)¡sin(θ) - Ο¡m¡g¡L¡cos(θ) = Ο¡L¡cos(θ)(Fâ¡sin(θ) - m¡g)
Work done on the block by friction = Ο¡L¡cos(θ)(Fâ¡sin(θ) - m¡g)
Learn more about motion on an inclined plane here:
https://brainly.com/question/14020955
