Áp dụng định luật II Newton tổng quát cho vật trượt ngang có ma sát ta có:

\(\begin{array}{*{20}{l}} {{F_1}\; = {\rm{ }}15N{\rm{ }} = {\rm{ }}{F_{ms1}}\; = {\rm{ }}\mu .mg{\rm{ }}\left( 1 \right)}\\ {{F_2}\; = {\rm{ }}60N{\rm{ }} = {\rm{ }}{F_{ms2}}\; = {\rm{ }}\mu \left( {m + 25} \right).g{\rm{ }}\left( 2 \right)} \end{array}\)

Chia (2) cho (1) ta có: \(\frac{{F1}}{{F2}} = \frac{{\mu .mg}}{{\mu \left( {m + 25} \right).g}} = \frac{m}{{m + 25}} = \frac{{15}}{{60}} = \frac{1}{4}\)

\( \Rightarrow m = \frac{{25}}{3}\) Thay vào (1) Ta có hệ số ma sát trượt. là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaqaaaaa % aaaaWdbiaadAeapaWaaSbaaSqaa8qacaaIXaaapaqabaGcpeGaaiiO % aiabg2da9iaabccacaaIXaGaaGynaiaad6eacaqGGaGaeyypa0Jaae % iiaiaadAeapaWaaSbaaSqaa8qacaWGTbGaam4Caiaaigdaa8aabeaa % k8qacaGGGcGaeyypa0JaaeiiaiaadwlacaGGUaGaamyBaiaadEgacq % GHshI3caWG1cGaaeiiaiaab2dadaWcaaqaaiaabccacaWGgbWdamaa \begin{array}{l} {F_1}\; = {\rm{ }}15N{\rm{ }} = {\rm{ }}{F_{ms1}}\; = {\rm{ }}\mu .mg \Rightarrow \mu {\rm{ = }}\frac{{{\rm{ }}{F_{ms1}}\;}}{{mg}} = \frac{{15.3}}{{25.10}}{\rm{ = 0,18}}\\ \end{array}\)

Đáp án B