Answer to Question #270720 in Classical Mechanics for kyungsoo

Question

I drag a mass m=21 kg 

in a straight line, along a horizontal surface, a distance D=23 m 

. I drag it at constant speed v=0.90 m.s in a straight line using a horizontal force. The coefficients of friction are \mu_{s} = 1.2 μs​=1.2 and μk ​=1.1. How much work do I do? W =

W= _____  J 

.

Answer 1

As per the given question,

Given mass = 21 kg

Distance on the horizontal surface (d) = 23m

Speed = 0.90 m/sec

Coefficient of friction $\mu_s =1.2$

$\mu_k =1.1$

Friction force = $f_s=\mu mg=1.1\times21\times9.8=226.38N$

work due to friction =$f_s . d=226.38\times23=5206.74J$

Kinetic energy of the object = $\dfrac{mv^2}{2}=\dfrac{21\times 0.9^2}{2}=8.505J$

Hence net work = $8.505+226.38 =234.885J$