What is ax, the x-component of the object's acceleration? express your answer to two significant figures and include the appropr

Question

aniked [119]

3 years ago

12

What is ax, the x-component of the object's acceleration? express your answer to two significant figures and include the appropr

iate units?

Physics

2 answers:

aliina [53] · Answer 1 · 2022-06-17T09:11:47+03:00

<span>Answer: F = ma Fx = max WHERE: Fx = forces in the x direction m = mass ax = acceleration in the x direction If we look at the figure, there are only two forces in the x direction. The first force is the x component force 4.0 N acting to the right, denoted as +4.0 N. The second force is the 2.0 N force acting to the left, denoted as -2.0 N. Fx = max 4.0 N - 2.0 N = (7.8 kg)*ax 2.0 N = (7.8 kg)*ax ax = 0.25641 m/s^2 Because the question is asking you to write it to two significant figures, ==>ax = 0.26 m/s^2</span>

o-na [289] · Answer 2 · 2022-06-17T11:26:02+03:00

The X-component of acceleration of the body is the rate of change of velocity of the body in the direction of X-axis.

Further Explanation:

Consider a body is moving with a uniform acceleration in $x-y\text{ plane}$ .

Here, $x$ represents the X-axis and $y$ represents the Y-axis of the plane.

The rate of change of position of the body on x-y plane can be specified by the rate of change of position in the direction of x-axis and y-axis.

$\vec r=\left( {x\hat i+y\hat j} \right){\text{ m}}$

Here, $\vec r$ is the position vector of the body, $x$ is the distance measure from origin in X-direction and $y$ is the distance measure from origin in Y-direction.

Similarly, the rate of change of velocity of the body can be specified by rate of change of velocity in direction of X-axis and Y-axis.

$\dfrac{{d\vec v}}{{dt}}=\left( {\left( {\dfrac{{d{v_x}}}{{dt}}} \right)\hat i + \left( {\dfrac{{d{v_y}}}{{dt}}} \right)\hat j} \right){{{{\text{ m}}} \mathord{\left/ {\vphantom {{{\text{ m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}^2}$ …… (1)

Here, $\dfrac{{d\vec v}}{{dt}}$ is the rate of change of velocity of body, $\left( {\dfrac{{d{v_x}}}{{dt}}} \right)$ is the rate of change of velocity in direction of X-axis and $\left({\dfrac{{d{v_y}}}{{dt}}}\right)$ is the rate of change of velocity in direction of Y-axis.

From the definition of acceleration,

The rate of change of velocity is acceleration and rate of change of velocity in X-direction is the acceleration in X-direction and rate of change of velocity in Y-direction is the acceleration in Y-direction.

So, equation (1) can be written as,

$\boxed{a=\left( {{a_x}\hat i + {a_y}\hat j} \right){{{\text{ m}}} \mathord{\left/ {\vphantom {{{\text{ m}}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}}$

Here, $a$ represents the actual or resultant acceleration of the body, ${a_x}$ is the acceleration of the body in X-direction or rate of change of velocity in X-direction, ${a_y}$ is the acceleration of the body in Y-direction or rate of change of velocity in Y-direction.