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Chapter 8: Problem 37

A small block of mass \(2 \mathrm{~kg}\) is kept on a rough inclined surface of inclination \(\theta=30^{\circ}\) fixed in a lift. The lift goes up with a uniform speed of \(1 \mathrm{~ms}^{-1}\) and the block does not slide relative to the inclined surface. The work done by the force of friction on the block in a time interval of \(2 \mathrm{~s}\) is (1) Zero (2) \(9.8 \mathrm{~J}\) (3) 294 I (4) \(16.9 \mathrm{~J}\)

Short Answer

Expert verified
The work done by the force of friction is zero.

Step by step solution

01

Analyze the Motion of the Lift

The lift is moving at a constant speed of \(1 \mathrm{~m/s}\), which means there's no acceleration. The only forces acting on the block from the inclined plane are gravity, the normal force, and friction. Since there's no relative movement between the block and the inclined surface, these forces are in equilibrium.
02

Determine Forces Acting on the Block

The gravitational force can be split into two components due to the inclined plane: parallel and perpendicular to the surface. The force parallel to the incline is \(mg\sin\theta\) and the normal force balances the perpendicular component \(mg\cos\theta\). The frictional force acts opposite to the potential sliding direction but since there is no sliding, this force will provide just enough force to counteract the gravity's component parallel to the incline.
03

Calculate Frictional Force

In this setup, since the block doesn't slide, the frictional force is equal to the component of the gravitational force parallel to the incline: \(f = mg\sin\theta\). Substituting the given values, \(m = 2 \mathrm{~kg}\), \(g = 9.8 \mathrm{~m/s^2}\), and \(\theta = 30^{\circ}\), we have: \(f = 2 \times 9.8 \times \sin 30^{\circ} = 9.8 \mathrm{~N}\).
04

Calculate Work Done by Friction

Work is calculated as force multiplied by displacement in the direction of the force. Since the lift moves the block without relative motion on the surface, the displacement with respect to the incline adds no work done by the force of friction. Therefore, the work done by friction in the time interval \((t = 2 \mathrm{~s})\) is zero, because the frictional force acts along the incline with no displacement in the direction of the force, due to its static nature.

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Key Concepts

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

Friction on Inclined Planes
When a block is on an inclined plane, like in the exercise, we deal with friction that prevents the block from sliding down. This frictional force is called static friction because the block does not move. Friction on inclined planes is mainly governed by:

  • The angle of inclination: This influences how gravity is split into components acting along and perpendicular to the plane.
  • The gravitational component parallel to the plane: It tries to pull the block down the slope.
  • The frictional force: This acts upward along the incline, opposing potential motion.
In the exercise, the incline is at an angle of 30 degrees. This angle is crucial as it determines calculations for how strong the friction must be to prevent sliding. The static friction matches exactly with the gravitational component along the plane causing the block to stay at rest.
Work and Energy
Work and energy are fundamental concepts in physics that describe the effort needed to move objects and the resulting motion or lack thereof. In the context of an inclined plane:
  • Work is defined as the force applied over a distance. The formula is: \ \( \text{Work} = \text{Force} \times \text{Distance} \times \cos(\phi) \ \), where \( \phi \) is the angle between force and direction of motion.
  • Energy comes into play when considering how forces like friction affect motion.
In the exercise, despite the movement of the block with the lift, no relative motion on the plane means no work is done by static friction. Even though the frictional force exists, it doesn't cause motion along its direction, leading to zero work.
Forces in Equilibrium
Forces acting on a body can be balanced, leading to equilibrium. When a system is in equilibrium:
  • The net force acting on the body is zero.
  • This implies no acceleration and, often, no movement relative to a surface, akin to the block on the lift's inclined plane.
In the case of the block on the inclined plane: - The component of gravitational force down the incline is counterbalanced by static friction. - The normal force balances the perpendicular gravitational component, keeping the block steady.
Thus, the net force is zero, ensuring the block doesn鈥檛 slide down the slope, demonstrating equilibrium, and resulting in zero work performed by the friction.

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Most popular questions from this chapter

A particle is projected vertically upwards with a speed of \(16 \mathrm{~m} \mathrm{~s}^{-1} .\) After some time, when it again passes through the point of projection, its speed is found to be \(8 \mathrm{~ms}^{-1}\). It is known that the work done by air resistance is same during upward and downward motion. Then the maximum height attained by the particle is (take \(g=10 \mathrm{~ms}^{-2}\) ) (1) \(8 \mathrm{~m}\) (2) \(4.8 \mathrm{~m}\) (3) \(17.6 \mathrm{~m}\) (4) \(12.8 \mathrm{~m}\)

Which of the following statements is correct? (1) Kinetic energy of a system can be changed without changing its momentum. (2) Kinetic energy of a system cannot be changed without changing its momentum. (3) Momentum of a system cannot be changed without changing its kinetic energy. (4) A system cannot have energy without having momentum.

A block of mass \(5.0 \mathrm{~kg}\) slides down from the top of an inclined plane of length \(3 \mathrm{~m}\). The first \(1 \mathrm{~m}\) of the plane smooth and the next \(2 \mathrm{~m}\) is rough. The block is releas from rest and again comes to rest at the bottom of the planee If the plane is inclined at \(30^{\circ}\) with the horizontal, find the coefficient of friction on the rough portion. (1) \(\frac{2}{\sqrt{3}}\) (2) \(\frac{\sqrt{3}}{2}\) (3) \(\frac{\sqrt{3}}{4}\) (4) \(\frac{\sqrt{3}}{5}\)

A man is standing on a plank which is placed on smooth horizontal surface. There is sufficient friction between feet of man and plank. Now man starts running over plank. The correct statements is/are (1) Work done by friction on man with respect to ground is negative (2) Work done by friction on man with respect to ground is positive (3) Work done by friction on plank with respect to ground is positive (4) Work done by friction on man with respect to plank is zero

One end of a spring of force constant \(k_{1}\) is attached to the ceiling of an elevator. A block of mass \(1.5 \mathrm{~kg}\) is attached to the other end. Another spring of force constant \(k_{2}\) is attached to the bottom of the mass and to the floor of the elevator as shown in the figure. At equilibrium, the deformation in both the springs is equal and is \(40 \mathrm{~cm}\). If the elevator moves with constant acceleration upward, the additional deformation in both the springs is \(8 \mathrm{~cm}\). Find the elevator's acceleration \(\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)\).
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