A block of mass m slides down a rough inclined plane of inclination with a constant downward acceleration a . The magnitude of the total contact force exerted by the plane on the block is :

Step-by-step solution

Let N be the normal force and f be the frictional force exerted by the incline on the block. Resolving forces perpendicular to the incline, the block has no acceleration, so: N = mg Applying Newton's second law along the incline (downwards): mg - f = ma f = m(g - a) The total contact force is the vector sum of the normal force and the frictional force. Since they are perpendicular, its magnitude is: F_ contact = N^2 + f^2 F_ contact = (mg )^2 + m^2(g - a)^2 F_ contact = m g^2 ^2 + g^2 ^2 + a^2 - 2ag F_ contact = m