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Rama09
3 months ago
6

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?

Physics
1 answer:
Yuliya22 [3.3K]3 months ago
5 0

Answer:

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

Explanation:

The Free Body Diagram depicting the assembly consisting of the crate and rope, alongside the reference axis, is illustrated in the image provided. Following is a presentation of the equilibrium equations:

\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

Through some algebraic manipulation, the equations consolidate into a single 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 necessary to initiate the crate's acceleration from a stationary state 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

As indicated by T > T_{min}, the crate will experience acceleration due to tension. Thus, the crate will have an acceleration of:

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}}

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