An object, whose mass is 0.660 kg, is attached to a spring with a force constant of 132 N/m. The object rests upon a frictionles
s, horizontal surface (shown in the figure below).
An object labeled m is attached to the right end of a horizontal spring, and the left end of the spring is attached to a wall. The spring is stretched horizontally such that the object is displaced by a distance A to the right of its equilibrium position.
The object is pulled to the right a distance A = 0.120 m from its equilibrium position (the vertical dashed line) and held motionless. The object is then released from rest.
(a)
At the instant of release, what is the magnitude of the spring force (in N) acting upon the object?
N
(b)
At that very instant, what is the magnitude of the object's acceleration (in m/s2)?
m/s2
(c)
In what direction does the acceleration vector point at the instant of release?
Toward the equilibrium position (i.e., to the left in the figure).
Away from the equilibrium position (i.e., to the right in the figure).
The direction is not defined (i.e., the acceleration is zero).
You cannot tell without more information.
An object labeled m is attached to the right end of a horizontal spring, and the left end of the spring is attached to a wall. The spring is stretched horizontally such that the object is displaced by a distance A to the right of its equilibrium position.
The object is pulled to the right a distance A = 0.120 m from its equilibrium position (the vertical dashed line) and held motionless. The object is then released from rest.
(a)
At the instant of release, what is the magnitude of the spring force (in N) acting upon the object?
N
(b)
At that very instant, what is the magnitude of the object's acceleration (in m/s2)?
m/s2
(c)
In what direction does the acceleration vector point at the instant of release?
Toward the equilibrium position (i.e., to the left in the figure).
Away from the equilibrium position (i.e., to the right in the figure).
The direction is not defined (i.e., the acceleration is zero).
You cannot tell without more information.
1 answer:
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Answer:
a) t = 20 [s]
b) Can't land
Explanation:
To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.
a)
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = desacceleration = 5 [m/s^2]
t = time [s]
Note: the negative sign of the equation means that the aircraft slows down as it stops.
0 = 100 - 5*t
5*t = 100
t = 20 [s]
b)
Now we can find the distance using the following kinematics equation.
x - xo = distance [m]
x -xo = (0*20) + (0.5*5*20^2)
x - xo = 1000 [m]
1000 [m] = 1 [km]
And the runaway is 0.8 [km], therefore the jetplane needs 1 [km] to land. So the jetpalne can't land