Question

A cart 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?

Answer 1

`W=1432.27\ J`

Step-by-step explanation:

Given:

• mass of the cart,
  `m=22.2\ kg`

• inclination of rope above the horizontal,
  `\theta=27.5^(\circ)`

• displacement of cart,
  `s= 24\ m`

• coefficient of kinetic friction,
  `\mu_k=0.32`

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

`f=\mu_k.R` ................................................(1)

Here normal reaction R:

`F.sin\theta+R=m.g` ........................................(2)

Now since the body is moving with a constant velocity:

∴ Kinetic Frictional force = applied force

`f=F`

`\mu_k.R=F.cos\theta`

`R=(F.cos\theta)/(\mu_k)` ................................................(3)

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

`F.sin\theta+(F.cos\theta)/(\mu_k)=m.g`

`F=m.g/ (sin\theta+(cos\theta)/(\mu_k))`

`F=22.2* 9.8/ (sin 27.5+(cos27.5)/(0.32))`

`F=67.28\ N` is the force applied along the rope.

Now the work done by the rope:

`W=F.s\ cos\theta`

`W=67.28* 24* cos\ 27.5`

`W=1432.27\ J`

Answer 2

W = 1.432 KJ

Step-by-step 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 = (\mu mg)/(cos\theta + \mu sin\theta)`

`F = (0.32 * 22.2 * 9.8)/(cos 27.5^0+0.32 * sin27.5^0)`

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