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Chapter 6: Problem 95

Two block of masses \(7 \mathrm{~kg}\) and \(5 \mathrm{~kg}\) are placed in contact with each other on a smooth surface. If a force of \(6 \mathrm{~N}\) is applied on a heavier mass the force on the lighter mass is (a) \(3.5 \mathrm{~N}\) (b) \(2.5 \mathrm{~N}\) (c) \(7 \mathrm{~N}\) (d) \(5 \mathrm{~N}\)

Short Answer

Expert verified
The force on the lighter mass is 2.5 N (option (b)).

Step by step solution

01

Understand the system

The system consists of two blocks in contact on a smooth surface, where a force is applied to the heavier block. We need to find the force acting on the lighter block due to this applied force.
02

Identify relevant principle

Since the surface is smooth (frictionless), the force applied to the blocks causes them to accelerate together. We'll use Newton's second law, which states: \[ F = m \cdot a \] where \( F \) is the net force applied, \( m \) is the mass, and \( a \) is the acceleration.
03

Calculate total mass and acceleration

The total mass of the two blocks is the sum of their individual masses: \[ m = 7 \text{ kg} + 5 \text{ kg} = 12 \text{ kg} \]Using the force applied, \( F = 6 \text{ N} \), we calculate the acceleration:\[ a = \frac{F}{m} = \frac{6 \text{ N}}{12 \text{ kg}} = 0.5 \text{ m/s}^2 \]
04

Calculate force on lighter block

The force on the lighter block is due to the acceleration it experiences and can be calculated using Newton's second law:\[ F_{\text{light}} = m_{\text{light}} \cdot a = 5 \text{ kg} \cdot 0.5 \text{ m/s}^2 = 2.5 \text{ N} \]

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Key Concepts

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

Frictionless Surface
Imagine placing two blocks on an ice rink, without any friction. That's essentially what a frictionless surface is 鈥� it implies that no force is resisting the motion between the object and the surface. In physics problems, when we say 'frictionless,' it means that the objects can move without any hindrance in their path, akin to gliding smoothly.
On such surfaces, any external force is efficiently transferred into motion or acceleration of the objects. In our exercise, the two blocks on the smooth surface experience a force that results in their acceleration without any energy loss to friction.
  • No resistance: Friction doesn't counteract the impressed force.
  • Pure motion: Forces solely cause forward acceleration.
  • Ideal scenario: This is an idealization we often use to simplify calculations in physics.
Force and Acceleration
When you push or pull an object, it experiences a force, and according to Newton's Second Law, this causes it to accelerate. The relationship between force, mass, and acceleration is given by \( F = m \cdot a \). This equation tells us how much an object will accelerate when a certain force is applied.
For our exercise, a force of \(6 \text{ N}\) is applied to a system of blocks. The combined system accelerates as a result of this force.
Here are some key points about force and acceleration:
  • Proportionality: Acceleration is directly proportional to the net force applied.
  • Inverse of mass: Acceleration is inversely proportional to the total mass.
  • Unit consistency: In our exercise, the units must be consistent 鈥� newtons (\(\text{N}\)) for force and kilograms (\(\text{kg}\)) for mass.
On calculating, the acceleration is found to be \(0.5 \text{ m/s}^2\), distributing this acceleration between both blocks.
Combined Mass System
In the given exercise, we consider two blocks as a single unit or combined mass system. This means when calculating acceleration or force acting on either block, we think about them together first.
When we apply a force to this system, it affects the entire mass together.To solve such problems, follow these steps:
  • Add masses: Determine the total mass by summing up individual masses. Here, the total mass is \(12 \text{ kg}\).
  • Compute acceleration: Use the total mass to find the acceleration, given the external force.
  • Distribute force: Calculate the force experienced by each block based on its share of the total mass.
For the lighter block, the internal force due to the combined mass system gives it a push equivalent to \(2.5 \text{ N}\). This comes from its mass multiplied by the calculated acceleration.

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Most popular questions from this chapter

A body of mass \(M\) is kept on a rough horizontal surface (friction coefficient \(\mu\) ). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on the body is \(F\), where (a) \(F=\mathrm{Mg}\) (b) \(F=\mu \mathrm{MgF}\) \((c) M g \leq f \leq M g \sqrt{1+\mu^{2}}\) (d) \(M g \geq f \geq M g \sqrt{1+\mu^{2}}\)

When forces \(F_{1}, F_{2}, F_{3}\) are acting on a particle of mass \(m\) such that \(F_{2}\) and \(F_{3}\) are mutually perpendicular, then the particle remains stationary. If the force \(F_{1}\) is now removed, then the acceleration of the particle is (a) \(F / m\) (b) \(E F_{1} m F_{i}\) \((c)\left(F_{2}-F\right) / m\) (d) \(F / m\)

A body of mass \(0.05 \mathrm{~kg}\) is obgerved to fall with an acceleration of \(9.5 \mathrm{~ms}^{-2}\), The opposite force of air on the body is \(\left(g=9.8 \mathrm{~ms}^{-2}\right)\) (a) \(0.015 \mathrm{~N}\) (b) \(0.15 \mathrm{~N}\) (c) \(0.030 \mathrm{~N}\) (d) zero

A dise of mass \(10 \mathrm{~g}\) is kept floating horizontally in air by firing bullets, each of mass \(5 \mathrm{~g}\), with the same velocity at the same rate of 10 bullets per second. The bullets rebound with the same speed in positive direction. The velocity of each bullet at the time of impact is (a) \(196 \mathrm{cms}^{-1}\) (b) \(98 \mathrm{cms}^{-1}\) (c) \(49 \mathrm{cms}^{-1}\) (d) \(392 \mathrm{cms}^{-1}\)

A man weighing \(60 \mathrm{~kg}\) is standing on a trolley weighing \(240 \mathrm{~kg}\). The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley with a velocity of \(1 \mathrm{~ms}^{-1}\), then after \(4 \mathrm{~s}\), his displacement relative to the ground is (a) \(6 \mathrm{~m}\) (b) \(4.8 \mathrm{~m}\) (c) \(3.2 \mathrm{~m}\) (d) \(2.4 \mathrm{~m}\)
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