1 Answer

Ask a Question

Rafidheen · Answer 1 · 2023-03-01T18:49:18+0000

Answer:

Step-by-step explanation: A box that has a mass of 19.8kg is pushed across 16.4m on the floor with a force of 93.2N, the floor has 0.355 as the coefficient of kinetic friction:

Visual Illustration of the problem:

Net work done by roy:

$\begin{gathered} F_n=F_r+F_f\Rightarrow F_n=F_r+\mu N \\ \mu=0.355\Rightarrow\text{ Coefficient of kinatic friction} \\ N=(19.8\operatorname{kg})\cdot(9.8ms^(-2))\Rightarrow\text{ Normal force on the box} \\ \therefore\Rightarrow \\ F_n=(93.2N)-(0.355)\cdot(19.8\operatorname{kg}\cdot9.8ms^(-2)) \\ F_n=24.3158 \\ \therefore\Rightarrow \\ W_r=F_n\cdot d\Rightarrow W_r=(24.3158)\cdot(16.4m)=398.77912J \\ W_r=398.77912J \end{gathered}$

Roy pushes a box of mass 19.8 kilograms across a floor by exerting a horizontal force-example-1

0 Comments

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

0 Comments

Please log in or register to add a comment.

Other Questions