Question #270720

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 

.



1
Expert's answer
2021-11-24T12:43:17-0500

As per the given question,

Given mass = 21 kg

Distance on the horizontal surface (d) = 23m

Speed = 0.90 m/sec

Coefficient of friction μs=1.2\mu_s =1.2

μk=1.1\mu_k =1.1

Friction force = fs=μmg=1.1×21×9.8=226.38Nf_s=\mu mg=1.1\times21\times9.8=226.38N

work due to friction =fs.d=226.38×23=5206.74Jf_s . d=226.38\times23=5206.74J

Kinetic energy of the object = mv22=21×0.922=8.505J\dfrac{mv^2}{2}=\dfrac{21\times 0.9^2}{2}=8.505J

Hence net work = 8.505+226.38=234.885J8.505+226.38 =234.885J

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