A cube, whose mass is 0.660 kg, is attached to a spring with a force constant of 106 N/m. The cube rests upon a frictionless, horizontal surface (shown in the figure below).
The cube is pulled to the right a distance A = 0.140 m from its equilibrium position (the vertical dashed line) and held motionless. The cube is then released from rest.
(a) At the instant of release, what is the magnitude of the spring force (in N) acting upon the cube?
(b) At that very instant, what is the magnitude of the cube's acceleration (in m/s2)?
In what direction does the acceleration vector point at the instant of release?
- The direction is not defined (i. e., the acceleration is zero).
-Away from the equilibrium position (i. e., to the right in the figure).
- Toward the equilibrium position (i. e., to the left in the figure).
- You cannot tell without more information.
***Please refer to attached photo
F = - F x force acting on cube
F = -106 N/n * .14 m = -14.8 N direction opposite to force
a = F / m = -14.8 N / .66 kg = -22.5 m/s^2 acceleration of cube
direction is towards equilibrium point (opposite F)
the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j