The question is not complete and this is the complete question;
"A spring of negligible mass has a force constant k = 1500 N/m. You place the spring vertically with one end on the floor. You then drop a book of mass 1.20 kg onto it from a height of 0.900 m above the top of the spring. Find the maximum distance the spring will be compressed. Take the free fall acceleration to be 9.80 m/s².
Answer:
Maximum distance = 12.69cm
Step-by-step explanation:
Using the Law of Conservation of Energy and knowing that the energy of the book that transfers
to the spring is the change in the gravitational potential energy of the book, we can say:
PEspring (initial) + PEgravitational (initial) =
PE spring(final) + PE gravitational(final)
From the question, we can see that, Initially the spring is uncompressed and so PE spring(initial) = 0J.
Thus, we get:
PE spring(final) = PE gravitational(initial) − PE gravitational(final)
Since, PE spring(final) is the final potential energy that is stored in the spring after dropping the book on it, then, PE gravitational(initial)
is the initial gravitational energy of the book at the height ℎ, while PE gravitational(final) is the final gravitational energy of the book
when the spring is compressed by the maximum distance ∆.
Hence, we can write that;
(k/2)(∆x)² = mgh − mg(−∆x)
So,
(k/2)(∆x)² - mg(∆x) - mgh= 0
So, k = 1500; m=1.2kg; g=9.8m/s²; h=0.9m
Thus,
(1500/2)(∆x)² - (1.2x9.8)(∆x) - (1.2 x 9.81 x 0.9) = 0
750(∆x)² - 11.76(∆x) - 10.584 = 0
So this is now a quadratic equation, thus;
Using quadratic formula: x = - b
(∆x) = -b ± √[(b²— 4ac)/2a]
So, (∆x) = -(-11.76) + √[(11.76²— 4(750)(-10.584))/(2x750)]
Or (∆x) = -(-11.76) - √[(11.76²— 4(750)(-10.584))/(2x750)]
Solving for the roots, we have;
(∆x) = - 0.1112 or 0.1269
The maximum distance can’t be negative, so the correct answer will be the positive one.
So (∆x) = 0.1269m or 12.69cm
