A spring whose spring constant is 450 N/m is compressed 0.800 m. What speed (in meters/second) can it give to a 0.500 kg ball when released?
1/2 m v^2=1/2 k x^2
v=sqrt(kx^2/m)
A 550 N crate rests on the floor. How much work in joules is required to move it at constant speed 7.0 m along the floor, against a friction force of 150 N?
To find the speed the spring can give to the ball, we need to apply the principle of conservation of energy.
The potential energy stored in the spring when it is compressed is given by:
Potential energy = (1/2) * k * x^2
Where:
k is the spring constant (450 N/m)
x is the distance the spring is compressed (0.800 m)
Potential energy = (1/2) * 450 N/m * (0.800 m)^2
Potential energy = 1/2 * 450 N/m * 0.640 m^2
Potential energy = 144 J
According to the principle of conservation of energy, this potential energy will convert into kinetic energy as the spring releases, given by:
Kinetic energy = (1/2) * m * v^2
Where:
m is the mass of the ball (0.500 kg)
v is the speed of the ball
Since the potential energy and kinetic energy are equal, we can equate them:
144 J = (1/2) * 0.500 kg * v^2
Simplifying the equation:
288 = v^2
v^2 = 288
v ≈ 16.97 m/s
Therefore, the speed the spring can give to the ball when released is approximately 16.97 m/s.