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A body of mass \(2 k g\) is kept by pressing to a vertical wall by a force of \(100 \mathrm{~N}\). The friction between wall and body is o.3. Then the frictional force is equal to [Orissa JEE 20o3] (a) \(6 \mathrm{~N}\) (b) \(20 \mathrm{~N}\) (c) \(600 \mathrm{~N}\) (d) \(700 \mathrm{~N}\)

Short Answer

Expert verified
Therefore, the right answer is none of the options given. The calculated frictional force is 30 N.

Step by step solution

01

Identify and write down the given information.

From the problem statement, the following is given: Mass of the body(m) = 2 kg, Force pressing the body to the wall (F pressing) = 100 N, Coefficient of friction (µ) = 0.3
02

Recall the formula to calculate the frictional force.

The formula for calculating frictional force (Ffriction) when the body is not in motion is: Ffriction = µ * Normal force. Here, since the body is being pressed against the wall, the normal force is equal to the pressing force (F pressing).
03

Substitute the given values into the formula and calculate.

By entering the given values into the formula, we get: Ffriction = µ * F pressing = 0.3 * 100 = 30 N

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Force
When we talk about forces in physics, the normal force often comes into play. It's one of those forces that acts perpendicular to a surface where two objects are in contact. By its nature, the normal force is always directed away from the surface. For the body pressed against a vertical wall in our example, the normal force isn't much complicated. It's equivalent to the force used to press the body, which is given as 100 N in the problem. In simpler terms, if you press a book against the wall with a certain force, the wall pushes back on the book with an equal normal force. Keep in mind:
  • Normal force is equal but opposite to the component of the applied force perpendicular to the contact surface.
  • Without normal force, friction as we know it doesn't exist; they're closely related.
Coefficient of Friction
The coefficient of friction, denoted by the Greek letter \( \mu \), is very important to understand friction. It gives us an idea of how much frictional resistance two surfaces in contact can exert on each other. In this problem, the coefficient of friction is provided as 0.3. This value signifies a moderate amount of friction between the body and the wall.How does this help us? Well, the coefficient of friction is used with the normal force to find out the frictional force.Some key points:
  • The value of \( \mu \) is a ratio and has no units.
  • It can vary based on material, surface texture, and even temperature.
  • Higher \( \mu \) means greater frictional resistance; lower \( \mu \) means less.
Recognizing the coefficient helps in solving problems related to friction quickly by applying the formula: \( F_{friction} = \mu \times \text{Normal Force} \).
Forces on Inclined Planes
While the given problem doesn't directly deal with inclined planes, understanding these forces provides insights into how forces operate in physics. On an inclined plane, forces such as gravity, normal force, and friction all play roles, but they act at angles. This usually necessitates breaking them into components. The core idea is: - Forces parallel to the plane: These could be components of gravity or any applied force. - Forces perpendicular to the plane: These often include the normal force. Solving inclined plane problems often involves:
  • Finding the angle of the incline;
  • Breaking forces into components using trigonometry;
  • Applying Newton's laws to these components.
Understanding forces on inclined planes prepares one to tackle a wide range of physics problems, broadening your grasp of force interactions considerably.

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Most popular questions from this chapter

A \(2 \mathrm{~kg}\) mass starts from rest on an inclined smooth surface with inclination \(30^{\circ}\) and length \(2 \mathrm{~m}\). How much will it travel before coming to rest on a surface with coefficient of friction \(0.25\) (a) \(4 m\) (b) \(6 \mathrm{~m}\) (c) \(8 \mathrm{~m}\) (d) \(2 \mathrm{~m}\)

A body of mass \(5 \mathrm{~kg}\) rests on a rough horizontal surface of coefficient of friction o.2. The body is pulled through a distance of \(10 m\) by a horizontal force of \(25 N\). The kinetic energy acquired by it is \(\left(g=10 \mathrm{~ms}^{2}\right)\) [BAMCET (Med.) 2000] (a) \(330 \mathrm{~J}\) (b) \(150^{\circ} J\) (c) \(100 J\) (d) \(50 \mathrm{~J}\)

A body of mass \(10 \mathrm{~kg}\) is lying on a rough plane inclined at an angle of \(30^{\circ}\) to the horizontal and the coefficient of friction is o.5. The minimum force required to pull the body up the plane is (a) \(914 N\) (b) \(91.4 N\) (c) \(9.14 \mathrm{~N}\) (d) \(0.914 N\)

A body of \(5 \mathrm{~kg}\) weight kept on a rough inclined plane of angle \(30^{\circ}\) starts sliding with a constant velocity. Then the coefficient of friction is (assume \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (a) \(1 / \sqrt{3}\) (b) \(2 / \sqrt{3}\) (c) \(\sqrt{3}\) (d) \(2 \sqrt{3}\)

If a ladder weighing \(250 \mathrm{~N}\) is placed against a smooth vertical wall having coefficient of friction between it and floor is \(0.3\), then what is the maximum force of friction available at the point of contact between the ladder and the floor (a) \(75 \mathrm{~N}\) (b) \(50 \mathrm{~N}\) (c) \(35 \mathrm{~N}\) (d) \(25 \mathrm{~N}\)

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