The derived frequency equals 2.63 Hz. Explanation: For an object weighing 8.0 kg with a spring stretching 3.6 cm, calculations involving the spring constant and oscillation frequency lead to this specific oscillation rate.
We will use the equations of rotational kinematics,
(A)
(B)
Here, 
and 
denote the final and initial angular displacements, respectively, whereas 
and 
represent final and initial angular velocities, and 
is the angular acceleration.
We are provided with 
and 
.
By substituting these values into equation (A), we have
Now, using equation (B),
This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.
This can be determined using the principle of energy conservation. The ski lift begins with a velocity of v= 15.5 m/s, and all of its kinetic energy Ek converts into potential energy Ep, thus we set Ep equal to Ek.
Because Ek is given by (1/2)*m*v², where m denotes mass and v represents speed, while Ep equals m*g*h, where m is mass, g is 9.81 m/s², and h is height. Now:
Ek=Ep
(1/2)*m*v²=m*g*h, canceling out the mass,
(1/2)*v²=g*h, rearranging for height by dividing by g,
(1/2*g)*v²=h and substituting the values:
h=12.245 m. The hill's height rounded to the nearest tenth is h=12.25 m.
Utilizing the equation F = ma, where F represents the force applied by the machine, A denotes acceleration (equivalent to v/t, with v as velocity and t as time), and M symbolizes mass, we can calculate as follows: F = mv/t. Thus, F = (0.15kg) (30 – 0 m/s) / 0.5 s, resulting in F = 9 N.
