Answer:
i'm sorry i'm not a physics student
Explanation:
please send full question....
Answer:
c. 1600J
Explanation:
The loss in potential energy of the boy is given by:
![U=mg \Delta h](https://tex.z-dn.net/?f=U%3Dmg%20%5CDelta%20h)
where
m = 40 kg is the mass of the boy
g = 9.8 m/s^2 is the acceleration of gravity
is the total change in the height of the boy (4 metres + 2 cm due to the compression of the spring)
Substituting, we find
![\Delta U = (40 kg)(9.8 m/s^2)(4.02 m) = 1577 J \sim 1600 J](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20%2840%20kg%29%289.8%20m%2Fs%5E2%29%284.02%20m%29%20%3D%201577%20J%20%5Csim%201600%20J)
Answer:
v = 0
Explanation:
This problem can be solved by taking into account:
- The equation for the calculation of the period in a spring-masss system
( 1 )
- The equation for the velocity of a simple harmonic motion
( 2 )
where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block
Hence
![T = \sqrt{\frac{2 kg}{50 N/m}} = 0.2 s](https://tex.z-dn.net/?f=T%20%3D%20%5Csqrt%7B%5Cfrac%7B2%20kg%7D%7B50%20N%2Fm%7D%7D%20%3D%200.2%20s)
and by reeplacing it in ( 2 ):
![v = \frac{2\pi }{0.2s}(14cm)sin(\frac{2\pi }{0.2s}(0.9s)) = 140\pi sin(9\pi ) = 0](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B2%5Cpi%20%7D%7B0.2s%7D%2814cm%29sin%28%5Cfrac%7B2%5Cpi%20%7D%7B0.2s%7D%280.9s%29%29%20%3D%20140%5Cpi%20%20sin%289%5Cpi%20%29%20%3D%200)
In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.