A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from rest and gains a speed of 15 m/s after sliding 150 m. How much work is done against friction

Respuesta :

Answer:

[tex]1270.64\ \text{J}[/tex]

Explanation:

m = Mass of object = [tex]\dfrac{mg}{g}[/tex]

mg = Weight of object = 20 N

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

[tex]\theta[/tex] = Angle of slope = [tex]30^{\circ}[/tex]

f = Force of friction

fd = Work done against friction

The force balance of the system is

[tex]\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}[/tex]

The work done against friction is [tex]1270.64\ \text{J}[/tex].