Ask question

Ask question

All categories

3 years ago

12

A man attempts to push a 19.8 kg crate across a warehouse floor. He slowly increases the force until the crate starts to move at

a force of 34.8 N. He then maintains this 34.8 N force while the box accelerates at 0.357 m/s^2. What is the coefficient of static friction between the crate and the floor?

2 answers:

VARVARA [1.3K]3 years ago

7 0

Answer:

μ = 0.18

Explanation:

Let's use Newton's second Law, the coordinate system is horizontal and vertical

Before starting to move the box

Y axis

N-W = 0

N = W = mg

X axis

F -fr = 0

F = fr

The friction force has the formula

fr = μ N

fr = μ m g

At the limit point just before starting the movement

F = μ m g

μ = F / m g

calculate

μ = 34.8 / (19.8 9.8)

μ = 0.18

notka56 [123]3 years ago

3 0

Answer:

The coefficient of friction between the crate and the floor = 0.143

Explanation:

Frictional Force: This is the force that act between two surface in contact and tend to oppose their motion. it is measured in Newton (N)

F - F₁ = ma.................... Equation 1

Where F = Force of the crate, F₁ = frictional force, m = mass of the crate, a = acceleration of the crate

making F₁ the subject the equation 1

F₁ = F - ma .................... Equation 2

<em>Given: F = 34. 8 N, m = 19.8 kg. a = 0.357 m/s².</em>

Substituting these values into equation 2

F₁ = 34.8 - (19.8×0.357)

F₁ = 34.8 - 7.07

F₁ = 27.73 N.

F₁ = μR ............... equation 3

making μ the subject of formula in equation 3

μ = F₁/R.............. Equation 4

Where F₁ = Frictional Force, μ = coefficient of static friction, R = Normal reaction.

R = mg,

where g = 9.8 m/s², m = 19.8 kg

R = 19.8( 9.8) = 194.04 N,

R = 194.04 N, F₁ = 27.73 N

Substituting these values into equation 4

μ = 27.73/194.04

μ = 0.143.

therefore the coefficient of friction between the crate and the floor = 0.143

You might be interested in

A ball is launched from ground level and hits the ground again after an elapsed time of 4 seconds and after traveling a horizont

jeka94

Answer:

1)a. It is constant the whole time the ball is in free-fall.

2)b. = 14 m/s

3) e. = 19.6 m/s

Explanation:

1) given that the only force acting on the ball is gravity, gravity acts along the vertical axis. Since no other force acts on the ball then the horizontal velocity will remain constant all through the flight since there is no horizontal force acting on the ball.

2) speed = distance/time

horizontal distance = 56m

Time = 4 seconds

Speed = 56m/4s = 14m/s

3) acceleration due to gravity g = 9.8m/s^2

Initial vertical velocity = u

Final vertical velocity = v = -u

Using the law of motion;

v = u + at

a = acceleration = -g = -9.8m/s^2

t = time of flight = 4

Substituting the values;

-u = u - 4(9.8)

-2u = -4(9.8)

u = -4(9.8)/-2

u = 2(9.8) = 19.6 m/s

Initial vertical velocity = u = 19.6 m/s

3 0

3 years ago

The weather conditions of an area may be expected to change over a(n) _________ period of time.

jenyasd209 [6]

Answer:

Your answer should be 2. Short

Explanation:

4 0

3 years ago

A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0

3 years ago

PLEASE HELP WITH GIVE BRAINLIEST!!! ❤️

sasho [114]

The force that the book exerts on the table is a normal force, not a weight force. (The book's weight doesn't act on the table, it acts on the book.) It's equal in magnitude to the weight of the book, again, because of the first law.

6 0

2 years ago

Q5- Knowing now that your mood might affect your memory, what strategies could you employ to assist your memories and make sure

Tema [17]

You can try recalling and rehearsing information about the memory.

4 0

2 years ago