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Chapter 3: Problem 331

Force of \(5 \mathrm{~N}\) acts on a body of weight \(9.8 \mathrm{~N}\). what is the acceleration produced in \(\mathrm{ms}^{-2}\). (A) \(49.00\) (B) \(5.00\) (C) \(1.46\) (D) \(0.51\)

Short Answer

Expert verified
(B) 5.00

Step by step solution

01

Convert weight to mass

In order to determine the mass of the object, we need to convert the weight from newtons (N) to mass in kilograms (kg). Since weight is the force of gravity acting on an object (mass x acceleration due to gravity), we can write: Weight = mass × gravitational acceleration So, we can find the mass by: Mass = Weight / gravitational acceleration Where: Weight = 9.8 N Gravitational acceleration ≈ 9.8 m/s² Inserting the values, we get: Mass (in kg) = \( \frac {9.8 \mathrm{~N}}{9.8 \mathrm {~m/s}^2} = 1 \mathrm{~kg}\)
02

Calculate the acceleration using Newton's second law

Now, we will apply Newton's second law of motion to calculate the acceleration produced from the given force. The formula for Newton's second law is: Force = Mass × Acceleration Rearranging the formula to find the acceleration: Acceleration = \( \frac{Force}{Mass}\) Where: Force = 5 N Mass = 1 kg Substitute the values: Acceleration = \( \frac{5 \mathrm{~N}}{1 \mathrm{~kg}} = 5 \mathrm{~m/s^2}\)
03

Find the answer in the given options

The calculated acceleration is 5 m/s², which corresponds to option (B) from the given options. So, the correct answer is: (B) 5.00

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Most popular questions from this chapter

A vehicle of \(100 \mathrm{~kg}\) is moving with a velocity of $5(\mathrm{~m} / \mathrm{s})\(. To stop it in \)(1 / 10) \mathrm{sec}$, the required force in opposite direction is \(\mathrm{N}\) (A) 50 (B) 500 (C) 5000 (D) 1000

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A car turns a corner on a slippery road at a constant speed of $10 \mathrm{~m} / \mathrm{s}\(. If the coefficient of friction is \)0.5$, the minimum radius of the arc at which the car turns is meter. (A) 20 (B) 10 (C) 5 (D) 4

A \(7 \mathrm{~kg}\) object is subjected to two forces (in newton) \(\underline{F}_{1}=20 \mathrm{i}^{-}+30 \mathrm{j}^{-}\) and \(\underline{\mathrm{F}}_{2}=8 \mathrm{i}^{-}-5 \mathrm{j}\) `The magnitude of resulting acceleration in \(\mathrm{ms}^{-2}\) will be (A) 5 (B) 4 (C) 3 (D) 2

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