Question

A cart loaded with bricks has a total mass of 28.3 kg and is pulled at constant speed by a rope. The rope is inclined at 23.1◦ above the horizontal and the cart moves 28.7 m on a horizontal floor. The coefficient of kinetic friction between ground and cart is 0.2. The acceleration of gravity is 9.8 m/s2. How much work is done on the cart by the rope? Answer in units of kJ

Answer 1

Step-by-step explanation:

Given

mass of cart with bricks
`m=28.3\ kg`

inclination
`\theta =23.1^(\circ)`

distance moved
`s=28.7\ m`

coefficient of kinetic friction
`\mu _k=0.2`

it is given that cart is moving with constant velocity so net force on it is Zero

if T is the tension in rope

Net normal Reaction

`N=mg-T\sin \theta`

Friction force
`f_r=\mu _k\cdot N`

`f_r=\mu _k(mg-T\sin \theta )`

Horizontal force
`=T\cos \theta`

therefore

`T\cos \theta =\mu _k(mg-T\sin \theta )`

`T=(\mu _kmg)/(\cos \theta +\mu_k\sin \theta )`

`T=(55.468)/(0.9982)`

`T=55.56\ N`

work done
`W=T\cos \theta \cdot s`

`W=55.56\cos (23.1)\cdot 28.7`

`W=1466.93\ J`