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Fxrbfg · Answer 1 · 2023-02-22T04:24:54+0000

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Step-by-step explanation: We need to find the total work that an object weighing 8kg does due to frictional force, the coefficient of kinetic friction is 0.4 and the object is pulled for 10 meters, the frictional force would be as follows:

$\begin{gathered} F_k=\mu N \\ \mu=0.4 \\ N=\text{ Normal force} \\ N=(8\operatorname{kg})*(10ms^(-2))=80N \\ \therefore\Rightarrow \\ F_k=(0.4)*(8\operatorname{kg})*(10ms^(-2))=32N \\ \\ \end{gathered}$

Therefore the total work done due to frictional forces is simply as follows:

$\begin{gathered} W=F_k\cdot d\cdot\cos (\theta)\rightarrow\theta=0 \\ \therefore\Rightarrow \\ W=(32N)*(10m)=320J \\ W=320J \end{gathered}$

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