Question

A body of mass 40kg is placed on an inclined plane with 60degrees as angle of inclination. if coefficient of friction = 0.6 , then the acceleration of the block is

Answer 1

$\textbf{\underline{\underline{As per Question :- }}}$

$\textbf{\underline{\underline{ Mass = 40 kg \theta = 60^{\circ} \mu = 0.6 }}}$

• Resolve mass in to its components.

There is only friction force acting on the body.

$\mathsf{f= ma }$

$\mathsf{\mu N =ma}$

$\mathsf{\mu mg Cos\theta= ma}$

$\textbf{ a = \mu g Cos60^{\circ} }$

$\textbf{ a = 0.6 \times 10 \times 0.5 }$

$\textbf{ a = 3 m/s^2 }$

hence,

The acceleration of the block is

$\textbf{\underline{\underline{ a = 3m/s^2 }}}$

Answer 2

Answer:

Explanation:

Angle of repose is simply the angle of inclination for which friction is max. i.e. u=tan(theta). If angle of inclination is less than the max. angle then friction will b more as tan theta has nw bcm small due to smaller angle, if angle of inclination is more than the theta for max friction then the block will slide down as friction wont b enogh to stop that block. In ur qstion its said that block has to b pushed down so angle of inclination is less than the angle or thetafor max. friction do option C is correct. Its kind of messy wht i wrote but read it twice or more n also frm ur physics book or notes u ll get it. Its very simple.Hope it helps!!