Question

A plane 5 meters in length is inclined at an angle of 37°, as shown in the figure. A block of weight 20 N is placed at the top of the plane and allowed to slide down.
The magnitude of the normal force exerted on the block by the plane is​

Answer 1

Answer:

16 N

Explanation:

See The Picture. Check It Out.

Answer 2

Normal Force by the plane is the reaction force by the plane due to the normal force block.

Thus, Normal force =. Component of Weight of block in the direction perpendicular to the inclined plane (Check Attachment for reference)

⇒ N = mgcosθ ......(1)

given,

1. weight mg = 20 N
2. θ = 37°

Substituting these values in equation (1) we get

N = 20 * cos37°

   = $20$ $*$ $\frac{4}{5}$

   = 16 N

⇒ The magnitude of the normal force exerted on the block by the plane is​ 16 N