(a) how far must the spring be compressed for 3.20 j of potential energy to be stored in it? (b) 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 0.800 m above the top of the spring. find the maximum distance the spring will be compressed.

a= 0.063 m

b = 0.116 m

Explanation:

First of all, we need the spring constant in order to solve this problem. You are not giving that data, but I will tell you how to solve this assuming a value of k, In this case, let's assume the value of k is 1600 N/m. (I solved an exercise like this before, using this value).

Now, we need to use the expressions to calculate the distance of the spring.

The elastic potential energy (Uel) is given with the following formula:

Uel = 1/2 kx²

Solving for x:

x = √2*Uel/k

Replacing the data in the above formula (And using the value of k os 1,600):

x = √2 * 3.2 / 1600

x = 0.063 m

b) For this part, we need to apply the work energy theorem which is:

K1 + Ugrav1 + Uel1 + Uo = K2 + Ugrav2 + Uel2

Since in this part, the exercise states that the book is dropped, we can say that the innitial and the end is 0, therefore, K1 = K2 = 0.

The spring at first is not compressed, so Uel1 = 0, and Uo which is the potential energy of other factors, is also 0, because there are no other force or factor here. Therefore, our theorem is resumed like this:

Ugrav1 = Uel2

The potential energy from gravity is given by:

Ug = mgy

And as the spring is placed vertically, we know the height which the book is dropped, so the distance y is:

y = x + h

And this value of x, is the one we need to solve. Replacing this in the theorem we have:

mg(h+x) = 1/2kx²

g would be 9.8 m/s²

Now, replacing the data:

1.2*9.8(0.8 + x) = 1/2*1600x²

Rearranging and solving for x we have:

1.2*9.8*2(0.8 + x) = 1600x²

18.82 + 23.52x = 1600x²

1600x² - 23.52x - 18.82 = 0

Now we need to solve for x, using the general formula:

x = - (-23.52) ± √(-23.52)² - 4 * 1600 * (-18.82) / 2*1600

x = 23.52 ± √553.19 + 120,448 / 3200

x = 23.52 ± 347.85 / 3200

x1 = 23.52 + 347.85 / 3200 = 0.116 m

x2 = 23.52 - 347.85 / 3200 = -0.101 m

Using the positive value, we have that the distance is 0.116 m.

d they tend to lose electrons and become cations.

explanation:

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.