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?
1 answer:
Answer:
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:
Through some algebraic manipulation, the equations consolidate into a single expression:
The minimum force necessary to initiate the crate's acceleration from a stationary state is:
As indicated by , the crate will experience acceleration due to tension. Thus, the crate will have an acceleration of:
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Answer:
a) τ = i ^ (y - z
) + j ^ (z Fₓ - x
) + k ^ (x
- y Fₓ)
b) τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m
c) α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²
Explanation:
a) The torque can be expressed as
τ = r x F
To tackle this equation, using the determinant approach is the most straightforward method
The resulting expression is
τ = i ^ (y - z
) + j ^ (z Fₓ - x
) + k ^ (x
- y Fₓ)
b) Now let's compute
τ = i ^ (0.075 1.4 -0.035 8.4) + j ^ (0.035 2.8 - 4.07 1.4) + k ^ (4.07 8.4 - 0.075 2.8)
τ = i ^ (- 0.189) + j ^ (-5.6) + k ^ (33,978)
τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m
c) To find angular acceleration, we use
τ = I α
α = τ / I
The moment of inertia being a scalar means that only the magnitude of each component changes, orientation remains constant.
α = (-0.189i^ -5.6 j^ + 33.978k^) / 241
α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²
The heat required to raise the temperature of a substance by 
is represented by
where m stands for the mass of the substance and
indicates the specific heat of the substance. In this situation, we possess 
and 
, the specific heat of water.
Consequently, we can ascertain the temperature rise:
Initially, the water's temperature was
, so the end temperature should be
Thus, the water is expected to be vapor by now.
However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as
where
denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of
(the temperature at which the phase change begins)
where m stands for the mass of the substance and
Consequently, we can ascertain the temperature rise
Initially, the water's temperature was
Thus, the water is expected to be vapor by now.
However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as
where
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of