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Chapter 3: Problem 370

A force of \(8 \mathrm{~N}\) acts on an object of mass \(5 \mathrm{~kg}\) in \(\mathrm{X}\) -direction and another force of \(6 \mathrm{~N}\) acts on it in \(\mathrm{Y}\) -direction. Hence, the magnitude of acceleration of object will be (A) \(1.5 \mathrm{~ms}^{-2}\) (B) \(2.0 \mathrm{~ms}^{-2}\) (C) \(2.5 \mathrm{~ms}^{-2}\) (D) \(3.5 \mathrm{~ms}^{-2}\)

Short Answer

Expert verified
The magnitude of the acceleration of the object is \(2~\frac{m}{s^{2}}\).

Step by step solution

01

Find the net force in X and Y directions

To find the net force in the X and Y directions, we can represent each force as a vector with components in the X and Y directions. The given forces on the object are: \(F_1 = 8~N\) in the X-direction and \(F_2 = 6~N\) in the Y-direction Since the forces are acting in perpendicular directions, their components in the X and Y directions can be represented as: \(F_1x = 8~N\), \(F_1y = 0~N\) and \(F_2x = 0~N\), \(F_2y = 6~N\) The net force in the X direction, \(F_{net\_x}\), is the sum of the forces in the X direction: \(F_{net\_x} = F_1x + F_2x = 8~N\) The net force in the Y direction, \(F_{net\_y}\), is the sum of the forces in the Y direction: \(F_{net\_y} = F_1y + F_2y = 6~N\)
02

Find the total net force

We can find the total net force, \(F_{net}\), acting on the object using the Pythagorean theorem for the X and Y components: \(F_{net} = \sqrt{F_{net\_x}^2 + F_{net\_y}^2} = \sqrt{(8~N)^2 + (6~N)^2} = 10~N\)
03

Use Newton's second law to find the acceleration

Newton's second law states that the net force acting on an object is equal to the product of the object's mass and its acceleration: \(F_{net} = m \times a\) We can rearrange the equation and solve for the acceleration: \(a = \frac{F_{net}}{m}\) Plugging in the given values: \(a = \frac{10~N}{5~kg} = 2~\frac{m}{s^2}\) The magnitude of the acceleration of the object is \(2~\frac{m}{s^{2}}\), which corresponds to option (B).

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Most popular questions from this chapter

A shell of mass \(200 \mathrm{~g}\) is ejected from a gun of mass $4 \mathrm{~kg}\( by an explosion that generates \)1.05 \mathrm{KJ}$ of energy. The initial velocity of the shell is (A) \(100 \mathrm{~m} / \mathrm{s}\) (B) \(80 \mathrm{~ms}^{-1}\) (C) \(40 \mathrm{~ms}^{-1}\) (D) \(120 \mathrm{~ms}^{-1}\)

Match the column \begin{tabular}{l|l} Column - I & Column - II \end{tabular} (a) Body line on a (p) is a self adjusting horizontal surface \(\quad\) force (b) Static friction (q) is a maximum value of static friction (c) Limiting friction (r) is then limiting friction (d) Dynamic friction (s) force of friction \(=0\) (A) $\mathrm{a}-\mathrm{s}, \mathrm{b}-\mathrm{p}, \mathrm{c}-\mathrm{q}, \mathrm{d}-\mathrm{r}$ (B) $\mathrm{a}-\mathrm{p}, \mathrm{b}-\mathrm{q}, \mathrm{c}-\mathrm{r}, \mathrm{d}-\mathrm{s}$ (C) $\mathrm{a}-\mathrm{s}, \mathrm{b}-\mathrm{r}, \mathrm{c}-\mathrm{q}, \mathrm{d}-\mathrm{p}$ (D) $\mathrm{a}-\mathrm{r}, \mathrm{b}-\mathrm{q}, \mathrm{c}-\mathrm{p}, \mathrm{d}-\mathrm{s}$

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