Answer:
g_x = 3.0 m / s^2
Explanation:
Given:
- Change in length of spring [email protected] = 22.6 cm
- Time taken for 11 oscillations t = 19.0 s
Find:
- The value of gravitational free fall g_x at plant X:
Solution:
- We will assume a simple harmonic motion of the mass for which Time is:
T = 2*pi*sqrt(k / m ) ...... 1
- Sum of forces in vertical direction @equilibrium is zero:
F_net = k*x - m*g_x = 0
(k / m) = (g_x / x) .... 2
- substitute Eq 2 into Eq 1:
2*pi / T = sqrt ( g_x / x )
g_x = (2*pi / T )^2 * x
- Evaluate g_x:
g_x = (2*pi / (19 / 11) )^2 * 0.226
g_x = 3.0 m / s^2
Explanation:
The given data is as follows.
Spring constant (k) = 78 N/m, 
Mass of block (m) = 0.50 kg
According to the formula of energy conservation,
mgh sin
h =
= 
= 0.64 m
Thus, we can conclude that the distance traveled by the block is 0.64 m.
Its called static friction.
Atmospheric pressure is molecular's heavy. So the total number of molecules decreases by going up the surface of the earth. Ok?