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

Answer:

$W=1432.27\ J$

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

$f=F$

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

$R=\frac{F.cos\theta}{\mu_k}$ ................................................(3)

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

$F.sin\theta+\frac{F.cos\theta}{\mu_k}=m.g$

$F=m.g\div (sin\theta+\frac{cos\theta}{\mu_k})$

$F=22.2\times 9.8\div (sin 27.5+\frac{cos27.5}{0.32})$

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

Now the work done by the rope:

$W=F.s\ cos\theta$

$W=67.28\times 24\times cos\ 27.5$

$W=1432.27\ J$

Answer 2

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

$F = \dfrac{\mu mg}{cos\theta + \mu sin\theta}$

$F = \dfrac{0.32 \times 22.2 \times 9.8}{cos 27.5^0+0.32 \times sin27.5^0}$

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