16. A 4.8-kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t = 0 s, the block has a displacement of -0.50 m, a velocity of -0.80 m/s, and an acceleration of +8.3 m/s2. The force constant of the spring is closest to
A. 56 N/m
B. 62 N/m
C. 73 N/m
D. 80 N/m
17. A 0.068 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 390 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = + 0.080 m and is released from rest. The block then executes SHM along the x-axis. The maximum elastic potential energy of the system is closest to
A. 1.1 J
B. 1.2 J
C. 1.4 J
D. 1.5 J
Oh spring, you're such a bouncy buddy! Let's tackle these questions with a touch of humor, shall we?
For question 16, let's find the force constant of the spring. We know that acceleration is equal to the negative of the displacement multiplied by the force constant. Plugging in the values, we get: 8.3 m/s^2 = -0.5 m * k. Solving for k, we get k = 16.6 N/m. But wait, we're not done clowning around yet! The closest answer choice is 16.6 N/m, so none of the options provided are correct. Looks like we stumped the multiple-choice test makers this time!
Moving on to question 17, we need to find the maximum elastic potential energy. Lucky for us, it's a frictionless surface, so we don't have to worry about any sneaky forces. The elastic potential energy is given by the formula U = (1/2) * k * x^2. Plugging in the values, we get U = (1/2) * 390 N/m * (0.080 m)^2 = 1.248 J. Well, it seems like the closest answer choice is 1.2 J, so let's go with that one!
And there you have it, my friend! Remember, even when dealing with serious physics questions, a little humor can brighten up the equation! Keep smiling and keep on learning!
To find the force constant of the spring in question 16, we need to use Hooke's Law:
F = -kx
where F is the force exerted by the spring, k is the force constant, and x is the displacement of the block. We also know that the acceleration of the block is given by:
a = F/m
where m is the mass of the block.
Let's use this information to find the force constant:
Given:
m = 4.8 kg
x = -0.50 m
a = 8.3 m/s^2
From Hooke's Law, we have:
ma = -kx
Substituting the given values:
4.8 kg * 8.3 m/s^2 = -k * (-0.50 m)
Simplifying:
39.84 N = 0.5k
k = 39.84 N / 0.5
k ≈ 79.68 N/m
Therefore, the force constant of the spring is closest to 80 N/m (option D).
Now, let's move on to question 17.
To find the maximum elastic potential energy of the system in question 17, we can use the formula:
U = (1/2)kx^2
where U is the elastic potential energy, k is the force constant of the spring, and x is the displacement of the block.
Given:
k = 390 N/m
x = 0.080 m
Substituting the given values into the formula:
U = (1/2)(390 N/m)(0.080 m)^2
U = 1/2 * 390 N/m * (0.0064 m^2)
U ≈ 1.2544 N/m * m^2
U ≈ 1.2544 J
Therefore, the maximum elastic potential energy of the system is closest to 1.2 J (option B).
To find the force constant of a spring in simple harmonic motion (SHM), we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
For question 16, we are given the displacement, velocity, and acceleration of the block at time t = 0s. Using these values, we can find the force acting on the block at that instant.
1. Start by calculating the restoring force acting on the block using Newton's second law: F = ma. In this case, the mass (m) of the block is given as 4.8 kg, and the acceleration (a) as +8.3 m/s².
F = (4.8 kg) * (8.3 m/s²) = 39.84 N
2. The restoring force of the spring is equal in magnitude but opposite in direction to the applied force. So, the force constant (k) of the spring can be calculated using the equation F = -kx, where x is the displacement. Since the block has a displacement of -0.50 m, the equation becomes:
39.84 N = (-k) * (-0.50 m)
3. Solve the equation for k:
k = 39.84 N / 0.50 m = 79.68 N/m
Rounding this value to the nearest whole number, we can conclude that the force constant of the spring is closest to 80 N/m. Therefore, the correct answer is option D.
For question 17, we are asked to find the maximum elastic potential energy of the system when the block is at its maximum displacement. In SHM, the maximum elastic potential energy occurs when the block is at the maximum displacement from its equilibrium position (amplitude).
1. The maximum displacement (x) of the block is given as +0.080 m, and the force constant (k) of the spring is 390 N/m.
2. The elastic potential energy (PE) of a spring is given by the equation: PE = (1/2)kx², where x is the displacement.
3. Calculate the maximum elastic potential energy:
PE = (1/2) * (390 N/m) * (0.080 m)²
PE = 1.248 J
Rounding this value to the nearest tenth, we can conclude that the maximum elastic potential energy of the system is closest to 1.2 J. Therefore, the correct answer is option B.