A horizontal 20.-newton force is applied to a 5.0-kilogram box to push it across a rough,horizontal floor at a constant velocity of 3.0 meters per second to the right.

Calculate the coefficient of kinetic friction between the box and the floor. [Show all work, including
the equation and substitution with units]

Answers

Answer 1

Answer: The formula to work out the coefficient of dynamic (kinetic) friction is F=uR where F is the force applied, u is the coefficient of dynamic friction, and R is the reaction force. This is based on the principle the object is moving at a constant velocity which is the case in this question.

Rearranging this we get u=F/R
Substituting in the values we get u=20/50=2/5=0.4

Therefore the coefficient of kinetic friction between the box and the floor is 0.4

Answer 2

Answer:

The coefficient of kinetic friction between the box and the floor will be $\boxed{0.4}$ .

Further Explanation:

Given:

The mass of the box is $5\,{\text{kg}}$ .

The force applied on the box is $20\,{\text{N}}$ .

The constant velocity at which the box moves is $3\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ .

Concept:

The box is pushed on the rough floor and due to this; the box will experience a friction force acting in the direction opposite to the motion of the box.

Since the box moves with a constant velocity of $3\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ , the acceleration of the box will be considered as zero.

From Newton’s second law of motion, as the acceleration of the box is zero, the net force acting on the box is zero.

$\boxed{{F_(net)}=m* a}$

The free-body diagram of the box moving on the rough floor is as shown in figure attached below.

From the above explanation, in order to make the net force zero, the friction acting on the box should be equal to the force applied on the box.

${F_(friction)}={F_(applied)}$

The friction force acting on the bxlock is:

$\begin{aligned}{F_(friction)}&=\mu N\n&=\mu mg\n\end{aligned}$

Substitute the value of friction force in the above expression.

$\begin{aligned}\mu mg&={F_(applied)}\n\mu\left({5*10}\right)&=20\n\mu &=\frac{{20}}{{50}}\n&=0.4\n\end{aligned}$

Thus, the coefficient of kinetic friction between the box and the floor will be $\boxed{0.4}$ .

Learn More:

1. A 30.0-kg box is being pulled across a carpeted floor by a horizontal force of 230 N brainly.com/question/7031524

2. Choose the 200 kg refrigerator. Set the applied force to 400 n (to the right). Be sure friction is turned off brainly.com/question/4033012

3. Which of the following is not a component of a lever brainly.com/question/1073452

Answer Details:

Grade: High School

Subject: Physics

Chapter: Friction

Keywords:

Horizontal 20N force, 5 kg box, push, a rough horizontal floor, constant velocity, friction force, coefficient of kinetic friction, box and the floor.

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