A spring is pulled to 10cm and held in place with a force of 500N. What is the spring constant of the spring
2 answers:
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Answer:
The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².
Explanation:
The centripetal acceleration is given by:
Where:
: is the tangential speed = 9.50 m/s
r: is the distance = 6.00 m
Hence, the centripetal acceleration is:
Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².
I hope it helps you!
To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.
According to Newton's second law we have to
where,
m= mass
g = gravitational acceleration
For the balance to break, there must be a mass M located at the right end.
We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.
In this way, applying the static equilibrium equations, we have to sum up torques at point B,
Regarding the forces we have,
Re-arrange to find M,
Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg
