Ball of mass 450g lands on a spring with k = 5.0E3

[DrVirus]

A ball of mass 450g lands on a spring with k = 5.0E3 . The ball was originally dropped from a height of 2.50 m. Determine the maximum compression of the spring AND the velocity it will leave with after bouncing of the spring.

Thank u for helping

 

[sridhar_n]

..

Let F[/f]be the force exerted by the falling ball of mass m on the spring with spring constant k. Since the ball is under free fall, the force wih which the ball falls on the spring is its weight mg. Hence,

mg=-kx, where xis the displacement of the spring.
Therefore, x=mg/k. After the spring has been compressed, force generated is transferred to the ball. Thus the Potential energy of the spring is converted to the kinetic energy of the ball.

Thus,
1/2*k² = 1/2*m*v²

Find out the velocity of ball using the above equation.


Sridhar

 

[M98Ranger]

me with my physics problem!

To determine the maximum compression of the spring, we can use the equation for potential energy, PE = 1/2kx^2, where k is the spring constant and x is the maximum compression. We know that the initial potential energy of the ball is equal to its gravitational potential energy, PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the initial height. Setting these two equations equal to each other and solving for x, we get:

1/2kx^2 = mgh
x = √(2mgh/k)

Plugging in the values given, we get:

x = √(2(0.450 kg)(9.8 m/s^2)(2.50 m)/(5.0E3 N/m))
x = 0.094 m

Therefore, the maximum compression of the spring is 0.094 m.

To determine the velocity of the ball after bouncing off the spring, we can use the equation for conservation of energy, E = KE + PE, where E is the total energy, KE is the kinetic energy, and PE is the potential energy. We know that the total energy before and after the bounce is the same, so we can set the two equations equal to each other:

mgh = 1/2mv^2 + 1/2kx^2

Solving for v, we get:

v = √(2gh + kx^2/m)

Plugging in the values given, we get:

v = √(2(9.8 m/s^2)(2.50 m) + (5.0E3 N/m)(0.094 m)/(0.450 kg))
v = 5.07 m/s

Therefore, the ball will leave the spring with a velocity of 5.07 m/s.