Two boxes are connected by a string. The first box has a mass of 4.2 kg and the
1 answer:
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11)
12)
13) 50.0 m/s
14) 41.6 m/s
Explanation:
11)
The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by
where
m is the mass of the object
g is the acceleration due to gravity
h is the height relative to the ground
Here in this problem, when the train is at the top, we have:
m = 8325 kg (mass of the train + riders)
(acceleration due to gravity)
h = 127 m (height of the train at the top)
Substituting,
12)
According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:
where
is the kinetic energy at the top
is the potential energy at the top
is the kinetic energy at the bottom
is the potential energy at the bottom
The kinetic energy is the energy due to motion; since the train is at rest at the top, we have
Also, at the bottom the height is zero, so the potential energy is zero
Therefore, we find:
13)
The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by
where
m is the mass of the object
v is the speed of the object
From question 12), we know that the kinetic energy of the train at the bottom is
We also know that the mass is
m = 8325 kg
Therefore, we can calculate the speed of the train at the bottom:
14)
At the top of the second hill, the total mechanical energy of the train is still conserved.
Therefore, we can write again:
where
is the kinetic energy at the top of the 1st hill
is the potential energy at the top of the 1st hill
is the kinetic energy at the top of the 2nd hill
is the potential energy at the top of the 2nd hill
From the previous questions, we know that
and
The height of the second hill is
h = 39 m
So we can also find the potential energy at the second hill:
So, the kinetic energy at the second hill is
And so, the speed is
Answer:
a. S(t)=350−1t
Explanation:
To determine the equation of motion you take into account the general form of motion with constant velocity:
( 1 )
So is the initial position from a specific reference frame. In this case is 350 m.
v is the speed of the motion, in this case is 1m/s. However, the motion is forward the zero point of the reference frame, hence, the speed is - 1m/s.
You replace the values of So and v in the equation ( 1 ) and you obtain:
Hence, the answer is:
a. S(t)=350−1t
- - - - - - - - - - - - - - - - - - - - -
Para determinar a equação do movimento, você leva em consideração a forma geral do movimento com velocidade constante:
(1)
Assim é a posição inicial de um quadro de referência específico. Neste caso, é de 350 m.
v é a velocidade do movimento, neste caso é de 1m / s. No entanto, o movimento é avançar o ponto zero do quadro de referência, portanto, a velocidade é de - 1m / s.
Você substitui os valores de So ev na equação (1) e obtém:
Portanto, a resposta é:
uma. S (t) = 350-1t, movimento retrógrado