/*! 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 66 A body of mass \(5 \mathrm{~kg}\... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 66

A body of mass \(5 \mathrm{~kg}\) is acted upon by two perpendicular forces \(8 \mathrm{~N}\) and \(6 \mathrm{~N}\). Give the magnitude of the acceleration of the body. (A) \(2 \mathrm{~m} / \mathrm{s}^{2}\) (B) \(4 \mathrm{~m} / \mathrm{s}^{2}\) (C) \(6 \mathrm{~m} / \mathrm{s}^{2}\) (D) \(8 \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The short answer is: \(a = 2 \mathrm{~m} / \mathrm{s}^{2}\) (A).

Step by step solution

01

Determine the net force

We have two perpendicular forces acting on the body: 8 N and 6 N. To find the net force, the magnitude of the resultant force, we can use the Pythagorean theorem. F_net = 鈭�(F1^2 + F2^2) where F1 = 8 N, F2 = 6 N, and F_net is the magnitude of the net force acting on the body.
02

Calculate the net force

Plug the values into the formula: F_net = 鈭�((8 N)^2 + (6 N)^2) = 鈭�(64 N^2 + 36 N^2) = 鈭�(100 N^2) = 10 N Now that we know the net force acting on the body, we can use Newton's second law to find the acceleration.
03

Apply Newton's second law

Newton's second law states that force equals mass times acceleration (F = ma). We are given the mass (m = 5 kg) and we just found the net force (F_net = 10 N). Now we can solve for the acceleration (a) by rearranging the equation: a = F_net / m
04

Calculate the acceleration

Plug in the known values into the equation: a = (10 N) / (5 kg) = 2 N/kg Acceleration has the units of meters per second squared (\(m/s^2\)), so the magnitude of the acceleration is 2 \(m/s^2\).
05

Identify the correct answer choice

Now that we have found the magnitude of the acceleration, we can identify the correct answer choice as (A) \(2 \mathrm{~m} / \mathrm{s}^{2}\).

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Most popular questions from this chapter

A particle of mass \(0.1 \mathrm{~kg}\) is whirled at the end of a string in a vertical circle of radius \(1.0 \mathrm{~m}\) at a constant speed of \(5 \mathrm{~m} / \mathrm{s}\). The tension in the string at the highest point of its path is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(0.5 \mathrm{~N}\) \(\begin{array}{lll}\text { (B) } 1.0 \mathrm{~N} & \text { (C) } 1.5 \mathrm{~N} & \text { (D) } 2.0 \mathrm{~N}\end{array}\)

A box of mass \(8 \mathrm{~kg}\) is placed on a rough inclined plane of inclination \(\theta\). Its downward motion can be prevented by applying an upward pull \(F\) and it can be made to slide upwards by applying a force \(2 F .\) The co-efficient of friction between the box and the inclined plane is (A) \(\frac{1}{3} \tan \theta\) (B) \(3 \tan \theta\) (C) \(\frac{1}{2} \tan \theta\) (D) \(2 \tan \theta\)

A block of mass \(m\) is placed on a surface with a vertical cross-section given by \(y=\frac{x^{3}}{6} .\) If the co-efficient of friction is \(0.5\), the maximum height above the ground at which the block can be placed without slipping is: (A) \(\frac{1}{6} m\) (B) \(\frac{2}{3} m\) (C) \(\frac{1}{3} m\) (D) \(\frac{1}{2} m\)

A block of \(10 \mathrm{~kg}\) is pulled by a constant speed on a rough horizontal surface by a force of \(19.6 \mathrm{~N}\). The co-efficient of friction is (A) \(0.1\) (B) \(0.2\) (C) \(0.3\) (D) \(0.4\)

A man of mass \(70 \mathrm{~kg}\) stands on a weighing scale in lift which is moving. Choose the correct statement. (A) Reading of weighing scale is \(700 \mathrm{~N}\) upwards with a uniform speed of \(10 \mathrm{~ms}^{-1}\). (B) Reading of weighing scale is \(700 \mathrm{~N}\) downwards with a uniform acceleration of \(5 \mathrm{~ms}^{-2}\) (C) Reading of weighing scale is \(700 \mathrm{~N}\) upwards with a uniform acceleration of \(5 \mathrm{~ms}^{-2}\) (D) Reading of weighing scale is \(700 \mathrm{~N}\) if the lift mechanism failed and it fall down freely under gravity.
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