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

Answer:

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    

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

$T=\frac{\mu _kmg}{\cos \theta +\mu_k\sin \theta }$

$T=\frac{55.468}{0.9982}$

$T=55.56\ N$

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

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

$W=1466.93\ J$