A 64kg driver gets into an empty taptap to start the day's work. The springs compress 1.9×10−2 m . What is the effective spring constant of the spring system in the taptap?
I have the same problem and that previous answer is wrong.
Why did the kangaroo bring a calculator to work? Because it wanted to compute its springiness!
To calculate the effective spring constant, we can use Hooke's Law: F = -kx, where F is the force, k is the spring constant, and x is the displacement of the spring.
In this case, the force is equal to the weight of the driver, which can be calculated as F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
So, F = (64 kg)(9.8 m/s^2) = 627.2 N.
Using Hooke's Law, we can then find the spring constant:
627.2 N = -k(1.9×10^-2 m).
Dividing both sides by -1.9×10^-2 m, we get:
k ≈ 33011.579 N/m.
Therefore, the effective spring constant of the spring system in the taptap is approximately 33011.579 N/m. Just don't let the springs "spring" a surprise on you while driving!
To find the effective spring constant of the spring system in the taptap, you can use Hooke's Law which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
The formula for Hooke's Law is:
F = k * x
Where:
F is the force exerted by the spring,
k is the spring constant, and
x is the displacement of the spring.
In this case, since the springs compress, the displacement is negative. Therefore, the equation becomes:
F = -k * x
To find the effective spring constant, we need to find the force exerted by the spring. The force can be calculated using the formula:
F = m * g
Where:
m is the mass of the driver, and
g is the acceleration due to gravity which is approximately 9.8 m/s².
Substituting the values, we get:
F = 64 kg * 9.8 m/s²
F = 627.2 N
Now, we have the force exerted by the spring, and we can rearrange Hooke's Law to calculate the spring constant:
k = -F / x
Substituting the values, we get:
k = -(-627.2 N) / (1.9 × 10^-2 m)
k = 33010.53 N/m
Therefore, the effective spring constant of the spring system in the taptap is approximately 33010.53 N/m.