In this problem, the formula we can use is:
v^2 = v0^2 – 2gd
where,
v = final velocity = 0 at the peak
v0 = 5 m/s
g = gravitational acceleration = 9.81 m/s^2
d = distance travelled
Calculating for d:
0 = 5^2 – 2(9.81) d
<span>d = 1.27 m</span>
-- In Roman Numerals, always ... 50 is larger than 40 .
-- In Algebra, whenever 'X' is less than ' 1 ' .
Assuming the temperature is kept constant during this change, you can use Boyle's Law to answer the question. The law relates the products of volume (V) and pressure (P) before and after:
![P_1V_1=P_2V_2](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2)
Given V1, P1, and V2:
![P_2 = \frac{P_1V_1}{V_2} = \frac{1.75atm\cdot 1.25l}{3.15l}=0.69atm](https://tex.z-dn.net/?f=P_2%20%3D%20%5Cfrac%7BP_1V_1%7D%7BV_2%7D%20%3D%20%5Cfrac%7B1.75atm%5Ccdot%201.25l%7D%7B3.15l%7D%3D0.69atm)
So, the pressure after the volume change, while the temperature is constant, will be 0.69 atmospheres. This is intuitive: the volume has increased with same amount of ideal gas, and so the pressure went down.
The force on the box is:
F = mgsin∅
If we multiply by this with the distance it traveled, we will know the work done by the box.
W = dmgsin∅
This work will be converted to elastic potential energy in the spring which is:
1/2 kx². Equating these and substituting values:
1/2 * 170 * x² = 4 * 13 * 9.81 * sin(30)
x = 1.73 m
The box's maximum speed will at the point right before contact with the spring, when the compression is 0.