Respuesta :
The acceleration of the box on the Carpeted floor is [tex]\fbox{\begin\ 0.67\,{{{\text{ m}}\mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}^2}\\\end{minispace}}[/tex] or [tex]\fbox{\begin\\\dfrac{2}{3}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\\end{minispace}}[/tex].
Further Explanation:
The friction force acting on the box is the force acting in the direction opposite to the motion of the box and therefore, the friction force opposes the relative motion of the box on the floor.
Given:
The mass of the box is [tex]30.0\,{\text{kg}}[/tex].
The force applied on the box is [tex]230\,{\text{N}}[/tex].
The friction force acting on the box is [tex]210\,{\text{N}}[/tex].
Concept:
The friction force acting on the box during the relative motion of the box on the floor reduces the net force acting on the box. The friction force acting on a body is of opposing nature and it opposes the motion of the body.
The net force acting on the box during motion is:
[tex]\fbox{\begin\\\bf{F_{{\text{net}}}} = F - f\\\end{minispace}}[/tex] …… (I)
Here, [tex]{F_{{\text{net}}}}[/tex] is the net force acting on the box, [tex]F[/tex] is the force applied on the box and [tex]f[/tex] is the friction force acting on the box during relative motion.
Substitute [tex]230\,{\text{N}}[/tex] for [tex]F[/tex] and [tex]210\,{\text{N}}[/tex] for [tex]f[/tex] in equation (I).
[tex]\begin{aligned}{F_{{\text{net}}}}&=230\,{\text{N}}-{\text{210}}\,{\text{N}}\\&=2{\text{0}}\,{\text{N}}\\\end{aligned}[/tex]
The acceleration of the box on the carpeted floor is due to the net force acting on the box and it is given as the ratio of net force and the mass of the box.
The acceleration of the box is given as:
[tex]\fbox{\begin\\\\\bf\ a =\dfrac{{{F_{{\text{net}}}}}}{m}\\\end{minispace}}[/tex] …… (II)
Here, [tex]a[/tex] is the acceleration of the box and [tex]m[/tex] is the mass of the box.
Substitute [tex]20\,{\text{N}}[/tex] for [tex]{F_{{\text{net}}}}[/tex] and [tex]30\,{\text{kg}}[/tex] for m in equation (II).
[tex]\begin{aligned} a &= \frac{{20\,{\text{N}}}}{{30\,{\text{kg}}}}\\&= \frac{2}{3}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}} \\&= 0.667\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}} \\ \end{aligned}[/tex]
Thus, the acceleration of the box on the carpeted floor is [tex]\fbox{\begin\ 0.67\,{{{\text{ m}}\mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}^2}\\\end{minispace}}[/tex] or [tex]\fbox{\begin\\\dfrac{2}{3}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\\end{minispace}}[/tex].
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Answer Details:
Grade: College
Subject: Physics
Chapter: Kinematics
Keywords:
Friction, acceleration, carpeted floor, force, relative motion, motion, net force, oppose, 2/3 m/s2, 2/3 m/s^2, 0.67 m/s^2, 0.67 m/s2, 2/3, 0.667 m/s^2, 0.667 m/s2.
Acceleration of the box is
[tex]\rm \dfrac{2}{3} \; m/sec^2[/tex]
Step by Step Solution :
Given :
Mass, m = 30 kg
Horizontal Force, F = 230 N
Friction Force, f = 210N
Calculation :
Refer the attached diagram for better understanding.
We know that,
[tex]\rm F_T= ma[/tex]
F - f = ma
230 - 210 = 30a
[tex]\rm a = \dfrac{2}{3}\;m/sec^2[/tex]
Therefore acceleration of the box is
[tex]\rm \dfrac{2}{3} \; m/sec^2[/tex]
For more information, refer the link given below
https://brainly.com/question/18754956?referrer=searchResults
