A block of mass m = 2.00 kg is attached to a spring of force constant k = 5.00 x102 N/m
that lies on a horizontal frictionless surface. The block is pulled to a position xi = 5.00 cm
to the right of equilibrium and released from rest. Find
a) the work required to stretch the spring and
b) the speed the block has as it passes through equilibrium
k = 5 * 10^2 Newtons / meter
x = 5 * 10^-2 meters
F = k x = 25 * 10^0 = 25 N
work in = integral k dx = (1/2) k x^2
= (1/2) * 5*10^2 * 25*10^-4 = (125/2) * 10^-2
= 1.25/2 = 0.625 Joules
then
(1/2) m v^2 = kinetic energy = 0.625 Joules
a) Well, pulling on a spring sounds like a stretchy situation. So, to calculate the work required to stretch the spring, we can use the formula:
Work = 0.5 * k * (xf^2 - xi^2)
Given that xi = 5.00 cm and k = 5.00 x 10^2 N/m, we can plug in the numbers and calculate the work!
b) Now, let's find the speed the block has as it passes through equilibrium. Since there is no friction involved, we can conserve mechanical energy.
The potential energy of the spring when stretched is equal to the kinetic energy of the block at equilibrium.
So, 0.5 * k * (xf^2 - xi^2) = 0.5 * m * v^2
We can solve for v by plugging in the values we know and doing some math!
Happy calculating! And remember, math might make you go "Spring, spring, spring!"
To find the work required to stretch the spring, we can use the formula for the potential energy of a spring:
1) The potential energy stored in a spring is given by the formula: U = (1/2)kx^2
where U is the potential energy, k is the force constant, and x is the displacement from equilibrium position.
In this case, the displacement of the block is 5.00 cm = 0.05 m, and the force constant is 5.00 x 10^2 N/m.
So, substituting the values into the equation, we get:
U = (1/2) * (5.00 x 10^2 N/m) * (0.05 m)^2
U = (1/2) * 250 N/m * 0.0025 m^2
U = 0.3125 J
Therefore, the work required to stretch the spring is 0.3125 Joules.
To find the speed of the block as it passes through equilibrium position, we can equate the potential energy at the starting point to the kinetic energy at the equilibrium position.
2) The potential energy at the starting point is equal to the work done to stretch the spring, which we have already calculated as 0.3125 J.
The kinetic energy at the equilibrium position is given by the formula: KE = (1/2)mv^2
Since the block is released from rest, its initial velocity is zero.
So, setting the potential energy equal to the kinetic energy:
0.3125 J = (1/2) * 2.00 kg * v^2
Simplifying the equation:
0.3125 J = 1.00 kg * v^2
Dividing both sides of the equation by 1.00 kg:
0.3125 J / 1.00 kg = v^2
Taking the square root of both sides of the equation:
√(0.3125 J / 1.00 kg) = v
v ≈ 0.559 m/s
Therefore, the speed of the block as it passes through equilibrium position is approximately 0.559 m/s.
To find the work required to stretch the spring, we can use the formula:
Work = (1/2) * k * (x_f^2 - x_i^2)
Where:
- Work is the work done to stretch the spring.
- k is the force constant of the spring.
- x_f is the final position of the block.
- x_i is the initial position of the block.
In this problem, the initial position of the block, x_i, is 5.00 cm to the right of equilibrium. Since we are measuring distances from equilibrium, this can be written as -0.05 m. The final position of the block, x_f, is at equilibrium, which corresponds to x_f = 0.
Plugging these values into the formula, we get:
Work = (1/2) * (5.00 x 10^2 N/m) * (0 - (-0.05)^2)
Work = (1/2) * (5.00 x 10^2 N/m) * (-0.05)^2
Work = (1/2) * (5.00 x 10^2 N/m) * 0.0025
Work = (1/2) * (5.00 x 10^2 N/m) * 0.0025
Work = 625 J
Therefore, the work required to stretch the spring is 625 Joules.
To find the speed of the block as it passes through equilibrium, we can use the principle of conservation of mechanical energy. Since there is no friction and the system is isolated, the total mechanical energy is conserved.
The mechanical energy is given by the sum of the kinetic energy and the potential energy:
Initial mechanical energy = Final mechanical energy
(1/2) * m * v_i^2 + (1/2) * k * x_i^2 = (1/2) * m * v_f^2 + (1/2) * k * x_f^2
Since the block is released from rest, the initial velocity, v_i, is 0. The final position, x_f, is at equilibrium, which corresponds to x_f = 0.
Plugging in the values, the equation becomes:
(1/2) * (2.00 kg) * (0^2) + (1/2) * (5.00 x 10^2 N/m) * (-0.05 m)^2 = (1/2) * (2.00 kg) * v_f^2 + (1/2) * (5.00 x 10^2 N/m) * 0^2
0 + (1/2) * (5.00 x 10^2 N/m) * 0.0025 m^2 = (1/2) * (2.00 kg) * v_f^2 + 0
0.625 J = (1/2) * (2.00 kg) * v_f^2
0.625 J = 1.00 kg * v_f^2
v_f^2 = 0.625 m^2/s^2
Taking the square root of both sides, we get:
v_f = 0.790 m/s
Therefore, the speed of the block as it passes through equilibrium is 0.790 m/s.