Respuesta :
Answer:
W = 1.432 KJ
Explanation:
given,
mass = 22.2 Kg
angle of the rope = 27.5°
distance on the ground = 24 m
kinetic friction= μ = 0.32
acceleration due to gravity, g = 9.8 m/s²
Work done = ?
W = F d cosθ
a = 0 because it is moving with constant speed
equating all the forces acting in x direction
F cosθ = F friction = μN
equating all the forces acting in y direction
F sinθ + N -mg =0
now,
N = mg - F sinθ
putting value of N
F cosθ = μ mg -μ F sinθ
F (cosθ + μsinθ ) = μ mg
[tex]F = \dfrac{\mu mg}{cos\theta + \mu sin\theta}[/tex]
[tex]F = \dfrac{0.32 \times 22.2 \times 9.8}{cos 27.5^0+0.32 \times sin27.5^0}[/tex]
F =67.28 N
now,
W=F d cosθ
W =67.28 x 24 x cos(27.5)
W =1432.27 J
W = 1.432 KJ
Answer:
[tex]W=1432.27\ J[/tex]
Explanation:
Given:
- mass of the cart, [tex]m=22.2\ kg[/tex]
- inclination of rope above the horizontal, [tex]\theta=27.5^{\circ}[/tex]
- displacement of cart, [tex]s= 24\ m[/tex]
- coefficient of kinetic friction, [tex]\mu_k=0.32 [/tex]
We find the force acting in the direction of the rope responsible for motion of the cart:
[tex]f=\mu_k.R[/tex] ................................................(1)
Here normal reaction R:
[tex]F.sin\theta+R=m.g[/tex] ........................................(2)
Now since the body is moving with a constant velocity:
∴ Kinetic Frictional force = applied force
[tex]f=F[/tex]
[tex]\mu_k.R=F.cos\theta[/tex]
[tex]R=\frac{F.cos\theta}{\mu_k}[/tex] ................................................(3)
Putting the value of R from eq. (3) into eq. (2)
[tex]F.sin\theta+\frac{F.cos\theta}{\mu_k}=m.g[/tex]
[tex]F=m.g\div (sin\theta+\frac{cos\theta}{\mu_k})[/tex]
[tex]F=22.2\times 9.8\div (sin 27.5+\frac{cos27.5}{0.32})[/tex]
[tex]F=67.28\ N[/tex] is the force applied along the rope.
Now the work done by the rope:
[tex]W=F.s\ cos\theta[/tex]
[tex]W=67.28\times 24\times cos\ 27.5[/tex]
[tex]W=1432.27\ J[/tex]
