Explain with method pls..
1 answer:
Answer:
See explanation below.
Explanation:
Given two blocks of masses M and m, And a force F acting on the block of mass M pulling it outside.
The two masses system can be considered as a whole, With F being the only external force acting on them.
Here the spring force has been ignored as while considering the two system as a whole, it will be accounted as an internal force for the system
Therefore, The acceleration of the system will be:
Now both the masses are moving with this same acceleration towards the direction of force .
Acceleration of mass m = a (as the system is moving as a whole)
Force on the mass m= mass times acceleration.
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Answer:
Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)
1) Todas las fuerzas están en la misma dirección.
Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:
F = 48N + 60N + 30N = 138N
Y por la segunda ley de Newton sabemos que:
F = m*a
fuerza igual a masa por aceleración.
Entonces la aceleración está dada por:
a = F/m = 138N/12kg = 11.5 m/s^2
2) Segundo caso, suponemos que F1 es opuesta a F2 y F3
En este caso, la fuerza neta será:
F = F2 + F3 - F1 = 60N + 30N - 48N = 42N
En este caso, la aceleración será:
a = 42N/12kg = 3.5 m/s^2
Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.