/*! 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 52 A force of \((5+3 x) \mathrm{N}\... [FREE SOLUTION] | 91Ó°ÊÓ

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A force of \((5+3 x) \mathrm{N}\) acting on a body of mass \(20 \mathrm{~kg}\) along the \(x\)-axis displaces it from \(x=2 \mathrm{~m}\) to \(x=6 \mathrm{~m}\). The work done by the force is (a) \(20 \mathrm{~J}\) (b) \(48 \mathrm{~J}\) (c) 68\(]\) (d) \(86 \mathrm{~J}\)

Short Answer

Expert verified
The work done is 68 J.

Step by step solution

01

Identify the Work Done Formula

The work done by a force is given by the integral of the force over the distance it acts. This requires using the integral formula: \[ W = \int F(x) \ dx \] where \( F(x) \) is the force as a function of \( x \).
02

Insert Function Into Integral

We plug the given force \( F(x) = 5+3x \) into the integral: \[ W = \int_{2}^{6} (5 + 3x) \, dx \]
03

Solve the Integral

Compute the integral \[\int (5 + 3x) \, dx = \int 5 \, dx + \int 3x \, dx = 5x + \frac{3x^2}{2} + C \]
04

Evaluate the Definite Integral

Evaluate the definite integral from \( x = 2 \) to \( x = 6 \) using the antiderivative:\[W = \left[5x + \frac{3x^2}{2}\right]_2^6 = \left(5(6) + \frac{3(6)^2}{2}\right) - \left(5(2) + \frac{3(2)^2}{2}\right).\]
05

Calculate the Result

Calculate the expression:\[W = \left(30 + 54\right) - \left(10 + 6\right) = 84 - 16 = 68 \, \text{J}.\]

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Key Concepts

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

Integral Calculus
Integral calculus is a powerful tool in mathematics used to find areas, volumes, and totals. Unlike differential calculus, which focuses on rates of change, integral calculus is all about summing up quantities. It uses the concept of a function being "integrated" over a particular region.

When you integrate, think of it as adding up lots of tiny pieces to find the whole. In our context, when we integrate a function representing force over a given distance, we end up finding the total work done. This is because work is essentially the sum of all the tiny force displacements across the distance.

In general terms, integrating helps determine overall quantities like:
  • The area under a curve.
  • The accumulated quantity of a rate of change.
  • Physical quantities in real-life scenarios, such as work, when force is applied.
This keeps integral calculus essential in physics and engineering, particularly in computing physical attributes such as energy and work.
Definite Integral
A definite integral computes the accumulation of values a function takes within a specific interval. Unlike an indefinite integral, which results in a general formula, a definite integral evaluates between two bounds, providing a specific numerical answer.

In the context of our problem, the definite integral is applied to the force function from the point where the body starts moving (at 2 meters) to where it stops (at 6 meters). The definite integral formula looks like:
  • \[ W = \int_{a}^{b} F(x) \, dx \]
This essentially means we're plugging into the formula the exact starting and ending points, labeled as 'a' and 'b', of our work computation. By substituting these limits, the integral makes the leap from a general function to the precise calculated work done by the force.

The result of this process is a finite, specific value—in this case, the work done over that span, helping physicists and engineers apply theoretical calculations to practical, real-world tasks.
Force as a Function of Distance
When considering a force as a function of distance, you're dealing with a situation where the force applied changes depending on the position. This is unlike constant force situations. In our example, the force is given by the function \( F(x) = 5 + 3x \). It illustrates a blend of constant force and varied force, increasing as the distance \( x \) increases.

Analyzing force as a function of distance can be complex, but integration simplifies this task. By creating a model of force dependent on distance, we can calculate the work done precisely, even if the force isn't constant over the entire path. This relationship is vital:
  • To predict how objects will interact with forces in changing conditions.
  • To compute energy required or released in various scenarios.
  • To apply practical solutions in engineering, physics, and beyond.
In physics, this concept helps ensure calculations and predictions accurately reflect reality in situations where forces aren't steady but are instead dynamically linked to position.

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Most popular questions from this chapter

A body of mass \(40 \mathrm{~kg}\) resting on a rough horizontal surface is subjected to a force \(P\) which is just enough to start the motion of the body. If \(\mu_{n}=0.5, \mu_{k}=0.4\), \(g=10 \mathrm{~ms}^{-2}\) and the force \(P\) is continuously applied on the body, then the acceleration of the body is (a) \(2 \mathrm{rro}\) (b) \(1 \mathrm{~ms}^{-2}\) (c) \(2 \mathrm{~ms}^{-2}\) (d) \(2.4 \mathrm{~ms}^{-2}\)

A body of mass \(2 \mathrm{~kg}\) travels according to law \(x(t)=p t+q t^{2}+r t^{3}\), where \(p=3 \mathrm{~ms}^{-1}, q=4 \mathrm{~ms}^{-1}\) and \(r=5 \mathrm{~ms}^{-3}\). The force acting on the body at \(t=2 \mathrm{~s}\) is [NCFRT Exemplat] (a) \(136 \mathrm{~N}\) (b) \(134 \mathrm{~N}\) (c) \(158 \mathrm{~N}\) (d) \(68 \mathrm{~N}\)

A cricket ball of mass 150 g collides straight with a bat with a velocity of \(10 \mathrm{~ms}^{-1}\). Batsman hits it straight back with a velocity of \(20 \mathrm{~ms}^{-1}\), If ball remains in contact with bat for \(0.18\), then average force exerted by the bat on the ball is (a) \(15 \mathrm{~N}\) (b) \(45 \mathrm{~N}\) (c) \(150 \mathrm{~N}\) (d) \(4.5 \mathrm{~N}\)

A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is \(0.25\), then the maximum fraction of the length of the chain that can hang over one edge of the table is (a) \(20 \%\) (b) \(25 \%_{6}\) (c) \(35 \%\) (d) \(15^{\circ 6}\)

A ball of mass \(0.2 \mathrm{~kg}\) is thrown vertically upwards by applying a force by hand. If the hand moves \(0.2 \mathrm{~m}\) while applying the force and the ball goes upto \(2 \mathrm{~m}\) height further, find the magnitude of the force [take \(\left.g=10 \mathrm{~ms}^{-2}\right]\) (a) \(16 \mathrm{~N}\) (b) \(20 \mathrm{~N}\) (c) \(22 \mathrm{~N}\) (d) \(4 \mathrm{~N}\)

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