Physics, 20.07.2019 13:40 artiomtyler007

To practice problem-solving strategy 5.2 dynamics problems. a 30.0-kg box is being pulled across a carpeted floor by a horizontal force of 230 n , against a friction force of 210 n . what is the acceleration of the box

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Answer from: verdolaga2017me

The acceleration of the box on the Carpeted floor is $\fbox{\begin\ 0.67\,{{{\text{ m}}\mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}^2}\\\end{minispace}}$ or $\fbox{\begin\\\dfrac{2}{3}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\\end{minispace}}$ .

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 $30.0\,{\text{kg}}$ .

The force applied on the box is $230\,{\text{N}}$ .

The friction force acting on the box is $210\,{\text{N}}$ .

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:

$\fbox{\begin\\\bf{F_{{\text{net}}}} = F - f\\\end{minispace}}$ …… (I)

Here, ${F_{{\text{net}}}}$ is the net force acting on the box, is the force applied on the box and is the friction force acting on the box during relative motion.

Substitute $230\,{\text{N}}$ for and $210\,{\text{N}}$ for in equation (I).

$\begin{aligned}{F_{{\text{net}}}}&=230\,{\text{N}}-{\text{210}}\,{\text{N}}\\&=2{\text{0}}\,{\text{N}}\\\end{aligned}$

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:

$\fbox{\begin\\\\\bf\ a =\dfrac{{{F_{{\text{net}}}}}}{m}\\\end{minispace}}$ …… (II)

Here, is the acceleration of the box and is the mass of the box.

Substitute $20\,{\text{N}}$ for ${F_{{\text{net}}}}$ and $30\,{\text{kg}}$ for m in equation (II).

$\begin{aligned} a &= \frac{{20\,{\text{N}}}}{{30\,{\text{kg}}}}\\&= \frac{2}{3}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}} \\&= 0.667\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}} \\ \end{aligned}$

Thus, the acceleration of the box on the carpeted floor is $\fbox{\begin\ 0.67\,{{{\text{ m}}\mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}^2}\\\end{minispace}}$ or $\fbox{\begin\\\dfrac{2}{3}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\\end{minispace}}$ .

Learn more:

1. Conservation of energy link

2. A ball falling under the acceleration due to gravity link

3. Translational kinetic energy of a molecule link

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.