The mass of a carbon atom is 2 x 10 ^ -26. What is the kinetic energy of a carbon atom moving with a speed of 500 m/s?
I got 2.5 x 10^-21 which is correct.
The second question states:
Two carbon atoms are joined by a spring-like carbon-carbon bond. The potential energy stored in the bond has the value calculated in part one if the bond is stretched 0.050nm. What is the bond's spring constant? Answer must be in N/m. Please help!
PE= 1/2 k x^2
solve for k
To calculate the bond's spring constant, you can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
Hooke's Law can be expressed as:
F = k * x
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the potential energy stored in the bond is equivalent to the kinetic energy of the carbon atom. Therefore, you can use the equation for kinetic energy:
KE = 0.5 * m * v^2
Where KE is the kinetic energy, m is the mass of the carbon atom, and v is the speed.
Given that the mass of a carbon atom is 2 x 10^-26 kg and the speed is 500 m/s, the kinetic energy can be calculated as:
KE = 0.5 * (2 x 10^-26 kg) * (500 m/s)^2
Now, you can equate the kinetic energy to the potential energy stored in the bond:
0.5 * (2 x 10^-26 kg) * (500 m/s)^2 = (1/2) * k * (0.050 x 10^-9 m)^2
Simplifying the equation:
(2 x 10^-26 kg) * (500 m/s)^2 = k * (0.050 x 10^-9 m)^2
Now, divide both sides of the equation by (0.050 x 10^-9 m)^2:
k = [(2 x 10^-26 kg) * (500 m/s)^2] / [(0.050 x 10^-9 m)^2]
Evaluating the equation:
k = [(2 x 10^-26 kg) * (500 m/s)^2] / [(0.050 x 10^-9 m)^2]
k ≈ 2.5 x 10^5 N/m
Therefore, the bond's spring constant is approximately 2.5 x 10^5 N/m.
To calculate the bond's spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position.
Hooke's Law can be written as:
F = -k * x
Where:
- F is the force exerted by the spring,
- k is the spring constant (which we want to find), and
- x is the displacement from the equilibrium position.
In this case, the potential energy stored in the bond is equal to the work done by the spring, which can be expressed as:
E = (1/2) * k * x^2
Given that the potential energy calculated in part one is 2.5 x 10^(-21) J and the displacement is 0.050 nm (which is 0.050 x 10^(-9) meters), we can substitute these values into the equation:
2.5 x 10^(-21) J = (1/2) * k * (0.050 x 10^(-9))^2
Simplifying:
5 x 10^(-21) J = (1/8) * k * (0.050 x 10^(-9))^2
Multiplying both sides by 8 to get rid of the fraction:
40 x 10^(-21) J = k * (0.050 x 10^(-9))^2
Now, let's calculate k:
k = (40 x 10^(-21) J) / (0.050 x 10^(-9))^2
Simplifying:
k = (40 x 10^(-21) J) / (0.050^2 x 10^(-9)^2)
k = (40 x 10^(-21) J) / (0.0025 x 10^(-18))
To divide these numbers, we can subtract their exponents:
k = (40 / 0.0025) x (10^(-21) / 10^(-18))
k = 16000 x 10^(-21 - (-18))
k = 16000 x 10^(-21 + 18)
k = 16000 x 10^(-3)
k = 16000 N/m
Therefore, the bond's spring constant is 16000 N/m.