Jeremy A.

asked • 06/18/21# Please Read the Question Below !

A block of mass 0.243 kg is placed on top of a light, vertical spring of force constant 4 975 N/m and pushed downward so that the spring is compressed by 0.097 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? (Round your answer to two decimal places.)

______m

## 1 Expert Answer

Amber M. answered • 06/19/21

B.S. Mathematics, M.S. Physics, Ph.D. candidate Physics

What you know:

m = 0.243 kg

k = 4975 N/m

x_{i} = 0.097 m

v_{i }= 0 m/s

x_{f} = 0 m

h_{i} = 0 m

v_{f }= 0 m/s

h_{f} = ?

This a conservation of energy question:

E_{i }= E_{f}

PE_{gi} + PE_{si} + KE_{i} = PE_{gf} + PE_{si }+ KE_{f}

mgh_{i} + 0.5kx_{i}^{2} + 0.5mv_{i}^{2} = mgh_{f} + 0.5kx_{i}^{2}_{ }+ 0.5mv_{f}^{2}

0.243(9.8)(0) + 0.5(4975)(0.097)^{2 }+ 0.5(0.243)(0)^2^{ }= 0.243(9.8)h_{f} + 0.5(4975)(0)^{2 }+ 0.5(0.243)(0)^2

23.40 = 2.38h_{f}

** **9.83 m = h_{f}

The height above the point of release is:

9.83 m - 0.097 m =** 9.73 m**

## Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem.

Most questions answered within 4 hours.

#### OR

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.

Rajai A.

06/18/21