Question

A block of mass 5 kg is placed on an inclined plane as shown in the figure. Force applied by incline plane on the block is (g = 9.8 m/s2)

Answer 1

force applied by inclined plane on the block is 49cosθ

explanation : you didn't attach figure. let inclination angle is θ.

weight of mass , W = mg = 5 × 9.8 = 49N [ vertically downward]

so, component of weight along plane is Wsinθ = 49sinθ N

and component of weight perpendicular to plane is Wcosθ = 49cosθ N

from Newton's 3rd law,

force applied by inclined plane on the block = component of weight perpendicular to plane = 49cosθ N

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Answer 2

$\huge\bold\purple{Answer:-}$

you didn't attach figure. let inclination angle is θ.

weight of mass , W = mg = 5 × 9.8 = 49N [ vertically downward]

so, component of weight along plane is Wsinθ = 49sinθ N

and component of weight perpendicular to plane is Wcosθ = 49cosθ N

from Newton's 3rd law,

force applied by inclined plane on the block = component of weight perpendicular to plane = 49cosθ N