Physics, 30.03.2020 18:04 barn01

Kayla, a fitness trainer, develops an exercise that involves pulling a heavy crate across a rough surface. The exercise involves using a rope to pull a crate of mass M = 45.0 kg along a carpeted track. Using her physics expertise, Kayla determines that the coefficients of kinetic and static friction between the crate and the carpet are μk = 0.410 and μs = 0.770, respectively. Next, Kayla has Ramon pull on the rope as hard as he can. If he pulls with a force of F = 325 N along the direction of the rope, and the rope makes the same angle of θ = 23.0 degrees with the floor, what is the acceleration of the box?

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Answer from: fatherbamboo

By pulling the heavy box crate along the rough surface, Ramon does work against friction and in moving the box crate.

The acceleration of the box is approximately 3.78 m/s².

Reasons:

Known parameters are;

Mass of the crate = 45.0 kg

Dynamic friction coefficient, $\mu_k$ = 0.410

Static friction coefficient, $\mu_s$ = 0.770

Force with which Ramon pulls the rope = 325 N

Angle the rope makes with the floor = 23.0°

Required:

Acceleration of the box

Solution:

The normal reaction = Weight of the box - vertical pull from Ramon

Therefore;

Normal reaction = 45.0 × 9.81 - 325 × sin(23.0°) ≈ 314.32 N

Force of Ramon pulling in the horizontal direction, Fₓ = 325 × cos(23.0°) ≈

299.16 N.

Friction force;

Static friction = 0.770 × 314.32 N ≈ 242.03 N

Kinetic , $\mu_k$ · N = 0.410 × 314.32 N ≈ 128.93 N

Given that the static friction force is less than Fₓ, the box will be put in

motion, and only the dynamic friction force applies.

Net force in the direction of motion = 325 × cos(23.0°) - 128.93 = 170.23

The net force in the direction of motion, F = 170.23 N

$Acceleration = \dfrac{Force}{Mass}$

$Acceleration \ of \ the \ box, \ a = \dfrac{170.23 \ N}{45.0 \, kg} \approx 3.78 \, m/s^2$

The acceleration of the box, α ≈ 3.78 m/s².

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Kayla, a fitness trainer, develops an exercise that involves pulling a heavy crate across a rough su

Answer from: riyah03

$a = 3.784\,\frac{m}{s^{2}}$

Explanation:

The Free Body Diagram of the system formed by the crate and the rope and the reference axis are presented below as an attached image. The equations of equilibrium are introduced hereafter:

$\Sigma F_{x} = T\cdot \cos \theta - \mu\cdot N = m\cdot a$

$\Sigma F_{y} = T\cdot \sin \theta + N - m\cdot g = 0$

After some algebraic handling, the system of equations is reduce to a sole expression:

$N = m\cdot g - T\cdot \sin \theta$

$T\cdot \cos \theta - \mu \cdot (m\cdot g - T\cdot \sin \theta) =m\cdot a$

$T\cdot (\cos\theta + \mu \cdot \sin \theta) - \mu \cdot m \cdot g = m\cdot a$

The minimum force to accelerate the crate from rest is:

$T_{min} = \frac{m\cdot \mu_{s}\cdot g}{\cos \theta + \mu_{s} \cdot \sin \theta}$

$T_{min} = \frac{(45\,kg)\cdot (0.770)\cdot (9.807\,\frac{m}{s^{2}} )}{\cos 23^{\textdegree}+(0.770)\cdot \sin 23^{\textdegree}}$

$T_{min} = 278.223\,N$

Since $T T_{min}$ , the crate will experiment an acceleration due to the tension exerted. The acceleration of the crate is:

$a = \frac{T}{m}\cdot (\cos \theta + \mu_{k}\cdot \sin \theta)-\mu_{k}\cdot g$

$a = \frac{325\,N}{45\,kg}\cdot [\cos 23^{\textdegree}+(0.410)\cdot \sin 23^{\textdegree}]-(0.410)\cdot (9.807\,\frac{m}{s^{2}} )$

$a = 3.784\,\frac{m}{s^{2}}$

Answer from: Quest

a longitudinal wave moves energy that is to the direction of the wave’s movement. - true

Answer from: Quest

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