Question

A plane surface is inclined at an angle of ${60}^0$ with the vertical. A body of mass 10 kg is uniformly accelerating, along the inclined plane surface. If the value of coefficient of friction ${\mu }_K$, between the body and the inclined surface is 0.2, calculate the resultant force of friction acting between the surfaces in contact. (Take $g=10m{s}^{-2}$)

• A. $17.32N$
• B. $5N$
• C. $7.5N$
• D. $10N$

Answer 1

Answer: (A)

$17.32N$

$F={\mu }_{K}mgsin60=0.2\times 10\times 10\times \frac{\sqrt{3}}{2}=17.32N$