MAXIMUM ELLONGATION IN SPRING WHEN THE MASS WITH SPRING IS THROWN WITH VELOCITY 4M FROM EQUILIBRIUM GIVEN THAT K=50 MASS OF BODY IS 5KG
To find the maximum elongation in a spring when a mass is thrown with a certain velocity, you need to consider the conservation of mechanical energy.
The formula for the maximum elongation of a spring is given by:
x = (m * v^2) / (2 * k)
Where:
- x is the maximum elongation of the spring
- m is the mass of the body attached to the spring (5 kg in this case)
- v is the velocity of the body (4 m/s in this case)
- k is the spring constant
Given that the spring constant (k) is 50 N/m, the mass of the body (m) is 5 kg and the velocity (v) is 4 m/s, you can substitute these values into the equation to find the maximum elongation of the spring.
x = (5 kg * (4 m/s)^2) / (2 * 50 N/m)
x = (5 * 16) / 100
x = 0.8 m
Therefore, the maximum elongation of the spring when the mass is thrown with a velocity of 4 m/s is 0.8 meters.