Acart loaded with bricks has a total mass of 22.2 kg and is pulled at constant speed by a rope. the rope is inclined at 27.5 ◦ degrees above the horizontal and the cart moves 24 m on a horizontal floor. the coefficient of kinetic friction between ground and cart is 0.32. the acceleration of gravity is 9.8 m/s 2 . how much work is done on the cart by the rope?

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

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

Explanation:

Given:

We find the force acting in the direction of the rope responsible for motion of the cart:

................................................(1)

Here normal reaction R:

........................................(2)

Now since the body is moving with a constant velocity:

∴ Kinetic Frictional force = applied force

................................................(3)

Putting the value of R from eq. (3) into eq. (2)

is the force applied along the rope.

Now the work done by the rope:

1/0.14nm = 10^10/1.4 a lot

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