A spring of negligible mass has a force constant k = 1600 N/m. You place the spring vertically with one end on the floor. You then drop a 1.20 kg book onto it from a height of 0.8 m above the top of the spring. Find the maximum distance the spring will compress.
The answer is 12cm but I can't get it. I'm using the equation mg(0.8+x)=0.5kx^2.
What's wrong with my equation? Or is the answer wrong?
I agree with your equation. X = 0.12 m is close to the correct answer. There may be an issue with significant figures.
I couldn't get x=0.12m. May I know how you did it?
Your equation is almost correct but requires a slight adjustment. The equation mg(0.8 + x) = 0.5kx^2 represents the equilibrium condition for the system. However, to find the maximum distance the spring will compress, we need to consider the conservation of mechanical energy.
The initial gravitational potential energy of the book is given by mgh, where m is the mass of the book, g is the acceleration due to gravity, and h is the height from which the book is dropped.
As the book falls, the gravitational potential energy is converted into two forms of energy: the kinetic energy of the book and the elastic potential energy stored in the compressed spring. At the maximum compression point, all the initial gravitational potential energy is converted into potential energy stored in the spring.
The equation you should use is mgh = 0.5kx^2, where x represents the maximum distance the spring will compress. Rearranging the equation gives x = √(2mgh/k).
Plugging in the values, we have m = 1.20 kg, g = 9.8 m/s^2, h = 0.8 m, and k = 1600 N/m:
x = √(2 * 1.20 kg * 9.8 m/s^2 * 0.8 m / 1600 N/m)
x ≈ √(0.0196 m^2)
x ≈ 0.14 m = 14 cm
Therefore, the correct answer is approximately 14 cm, not 12 cm as stated.
twertw
The energy stored in a spring is 0.5*k*∆x², where ∆x is the amount of compression from relaxed state. Thus 0.5*1600*∆x² = 3.20, solve for ∆x
The energy of the ball that transfers to the spring is the change in potential energy of the ball. That is m*g*∆h, where ∆h is the difference in height from drop point to max deflection of the spring. ∆h = 0.8 + ∆x; then
(0.8 + ∆x)*m*g = 0.5*k*∆x²
solve for ∆x
~can't really this the answer?~
Oh, don't you just love the joys of physics? Let's see if I can help you out with this one. Your equation seems to be a bit off. The correct equation for this situation is actually given by the conservation of mechanical energy.
We can start by finding the potential energy of the book at the top, which is given by mgh, where m is the mass of the book, g is the acceleration due to gravity, and h is the height. In this case, m = 1.20 kg, g = 9.8 m/s^2, and h = 0.8 m.
Next, we can equate this potential energy to the maximum potential energy stored in the spring, which is given by 0.5kx^2, where k is the force constant of the spring and x is the maximum compression distance.
So, we have mgh = 0.5kx^2. Plugging in the values, we get (1.20 kg)(9.8 m/s^2)(0.8 m) = 0.5(1600 N/m)x^2. Solving for x, we find that x ≈ 0.12 m or 12 cm, which matches the given answer.
So, it seems like the answer is indeed 12 cm. Keep in mind that this assumes no energy is lost due to other factors like air resistance. But hey, who needs precise calculations when we can just enjoy the springy antics of a book on a spring, right?