**Problem Statement:**A 37 N block rests on an inclined plane with a slope of 25°. What is the normal force acting on the block?**Explanation:**To solve this problem, you need to determine the normal force exerted on the block by the inclined plane. The normal force is the perpendicular force that the surface exerts on an object in contact with it. Given:- The weight of the block (\(W\)) is 37 N.- The angle of the incline (\(\theta\)) is 25°.To find the normal force (\(N\)), we consider the component of the weight that acts perpendicular to the inclined plane. This can be calculated using the formula:\[ N = W \cdot \cos(\theta) \]Where:- \(N\) is the normal force.- \(W\) is the weight of the block, 37 N.- \(\theta\) is the angle of the incline, 25°.- \(\cos(\theta)\) is the cosine of the angle.**Steps:**1. Calculate \(\cos(25°)\).2. Multiply the weight of the block by \(\cos(25°)\) to find the normal force.This calculation will give the amount of the normal force acting on the block due to the inclined plane.

Weight of object is Angle of inclination is The object rests on an inclined plane.Find:Normal force…

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A 37 N block rests on an inclined plane with a slope of 25°. What is the normal force acting on the block?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

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Torque

Torque is the rotational equivalent of linear force in physics and mechanics. Based on the course of study, it is also known as the moment of force, rotational force, or turning effect.

Question

**Problem Statement:**

A 37 N block rests on an inclined plane with a slope of 25°. What is the normal force acting on the block?

**Explanation:**

To solve this problem, you need to determine the normal force exerted on the block by the inclined plane. The normal force is the perpendicular force that the surface exerts on an object in contact with it.

Given:
- The weight of the block (\(W\)) is 37 N.
- The angle of the incline (\(\theta\)) is 25°.

To find the normal force (\(N\)), we consider the component of the weight that acts perpendicular to the inclined plane. This can be calculated using the formula:

\[ N = W \cdot \cos(\theta) \]

Where:
- \(N\) is the normal force.
- \(W\) is the weight of the block, 37 N.
- \(\theta\) is the angle of the incline, 25°.
- \(\cos(\theta)\) is the cosine of the angle.

**Steps:**
1. Calculate \(\cos(25°)\).
2. Multiply the weight of the block by \(\cos(25°)\) to find the normal force.

This calculation will give the amount of the normal force acting on the block due to the inclined plane.

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