/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A man moves on a straight horizo... [FREE SOLUTION] | 91Ó°ÊÓ

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A man moves on a straight horizontal road with a block of mass \(2 \mathrm{~kg}\) in his hand. If he covers a distance of \(40 \mathrm{~m}\) with an acceleration of \(0.5 \mathrm{~m} / \mathrm{s}^{2}\), find the work done by the man on the block during the motion.

Short Answer

Expert verified
The work done by the man on the block is 40 J.

Step by step solution

01

Identify the given values

We need to determine the work done on the block. The given values are: mass of the block, \( m = 2 \text{ kg} \); acceleration, \( a = 0.5 \text{ m/s}^2 \); and distance covered, \( s = 40 \text{ m} \).
02

Calculate the net force on the block

The force exerted can be calculated using Newton's second law: \( F = ma \). Substituting the given values: \( F = 2 \times 0.5 = 1 \text{ N} \).
03

Calculate the work done

Work done is given by the formula \( W = F \times s \), where \( F \) is the force and \( s \) is the distance. Substituting the values: \( W = 1 \times 40 = 40 \text{ J} \).
04

Verification

Ensure that units are consistent: Force is in Newtons, distance is in meters, and work done is in Joules. The calculation confirms these units are correctly applied in the formula.

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Key Concepts

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

Newton's Second Law
When we think about how forces work in relation to moving objects, it's vital to understand Newton's Second Law of Motion. This law is fundamental in physics and describes the relationship between an object's mass, the force applied to it, and the resulting acceleration. The second law can be simply expressed using the formula:
  • \( F = ma \)
In this equation, \( F \) represents the force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration of the object.
This law implies that the force needed to move an object is the product of the mass and the acceleration. It helps us understand why heavier objects require more force to move and why more force results in higher acceleration, as long as the mass stays the same.
Force Calculation
Calculating the force needed to accelerate an object is a practical application of Newton's Second Law. In the exercise, we used the formula \( F = ma \) to find the force that the man applied to the block.
  • The mass \( m \) of the block was \( 2 \text{ kg} \).
  • The acceleration \( a \) was \( 0.5 \text{ m/s}^2 \).
By multiplying these two values, we calculated that the force was \( 1 \text{ N} \).
This straightforward multiplication not only shows how much force is applied but also reveals how different factors affect the force exerted:
  • Increasing the mass or the acceleration will increase the force required.
  • Decreasing either will reduce the required force.
Understanding this can help in designing systems that involve movement, ensuring that they function efficiently.
Distance and Displacement
In physics, distance and displacement describe how far an object moves and in what direction. Distance refers to the total path length traveled by an object, regardless of direction, while displacement takes into account the object’s overall change in position.
In the provided exercise, the man carried a block over a distance of \( 40 \text{ m} \). This example uses distance because it measures the total ground covered, crucial for calculating the work done.
For calculating work done, the formula \( W = F \times s \) is used, where \( W \) is the work, \( F \) is the force, and \( s \) is the distance the force acts over. In our situation:
  • The force \( F \) was \( 1 \text{ N} \).
  • The distance \( s \) was \( 40 \text{ m} \).
These values produced a work done value of \( 40 \text{ J} \) (Joules), helping illustrate how forces translate into movement over a defined distance.

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Most popular questions from this chapter

A \(250 \mathrm{~g}\) block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if it is initially moving at a speed of 40 \(\mathrm{cm} / \mathrm{s}\). If the friction coefficient between the table and the block is \(0^{\circ} 1\), how far does the block move before coming to rest"?

A particle slides on the surface of a fixed smooth sphere starting from the topmost point. Find the angle rotated by the radius through the particle, when it leaves contact with the sphere.

A particle of mass \(m\) moves on a straight line with its velocity varying with the distance travelled according to the equation \(v=a \sqrt{x}\), where \(a\) is a constant. Find the total work done by all the forces during a displacement from \(x=0\) to \(x=d\).

The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity \(\sqrt{10 g l}\), where \(l\) is the length of the pendulum. Find the tension in the string when (a) the string is horizontal, (b) the bob is at its highest point and (c) the string makes an angle of \(60^{\circ}\) with the upward vertical.

A small block of mass \(200 \mathrm{~g}\) is kept at the top of a frictionless incline which is \(10 \mathrm{~m}\) long and \(3 \cdot 2 \mathrm{~m}\) high. How much work was required (a) to lift the block from the ground and put it at the top, (b) to slide the block up the incline? What will be the speed of the block when it reaches the ground, if (c) it falls off the incline and drops vertically on the ground (d) it slides down the incline ? Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\)

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