A 65.0 kg clown gets into a 1250 kg clown car that has a 55.0 kg clown in the driver's seat. The car gets 1.40 cm closer to the ground.
A) What is the k value of the car's spring?
B) If two other clowns (45.0 kg and 52.0 kg) get into the car, how much closer with the car get to the ground?
k=force/distanc=(65+1250+55)*9.8/1.40 N/m
how much closer= (45+52)(9.8)/kabove
the study guide says the answers are a) 4.5*10^4 N/m and b) 0.021 m but i still can't get those answers with your method
To solve this problem, we need to 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 equation for Hooke's Law is:
F = k * x
where F is the force applied, k is the spring constant, and x is the displacement from the equilibrium position.
A) To find the spring constant (k value), we need to use the equation:
k = F / x
where F is the weight of the clowns and x is the displacement of the car from the ground.
Let's calculate the force applied. The total weight applied to the spring is the sum of the weights of the clowns in the car.
Weight of the clown in the driver's seat = mass * acceleration due to gravity
= 55.0 kg * 9.8 m/s^2
= 539 N
Weight of the clown getting into the car = 65.0 kg * 9.8 m/s^2
= 637 N
Total force applied = 539 N + 637 N
= 1176 N
Now, let's calculate the displacement of the car from the ground. Given that the car gets 1.40 cm closer to the ground, we need to convert it to meters.
Displacement (x) = 1.40 cm * 0.01 m/cm
= 0.014 m
Finally, we can calculate the spring constant (k):
k = 1176 N / 0.014 m
= 84000 N/m
So, the k value of the car's spring is 84000 N/m.
B) To find how much closer the car will get to the ground when two other clowns get into the car, we need to calculate the total force applied and the new displacement of the car.
The weight of the clowns getting into the car is:
Weight of the first clown = 45.0 kg * 9.8 m/s^2
= 441 N
Weight of the second clown = 52.0 kg * 9.8 m/s^2
= 510 N
Total force applied when two other clowns get into the car = 1176 N + 441 N + 510 N
= 2127 N
Now, let's calculate the new displacement of the car. We know the spring constant remains the same.
Displacement (x) = ?
Using Hooke's Law equation:
F = k * x
We can rearrange the equation to solve for x:
x = F / k
x = 2127 N / 84000 N/m
x ≈ 0.025 m
Therefore, the car will get approximately 0.025 meters or 2.5 cm closer to the ground when two other clowns get into the car.