A ball with a mass of #280 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #42 (kg)/s^2# and was compressed by #5/3 m# when the ball was released. How high will the ball go?

Question

A ball with a mass of #280 g# is projected vertically by a spring loaded contraption.  The spring in the contraption has a spring constant of #42  (kg)/s^2# and was compressed by #5/3 m# when the ball was released.  How high will the ball go?

Layla Ahmed · Accepted Answer

#h=21,25" m"# #"given data:"# #Delta x=5/3 m# #K=42 (kg)/s^2# #m=280g=0,28 kg" mass of object"# #g=9,81 m/s^2# #h=?# #E_p=1/2*k*Delta x^2" Potential energy for compressed springs"# #E=m*g*h " potential energy for object raising from earth"# #E_p=E" conservation of energy"# #m*g*h=1/2*K*Delta x^2# #h=(K*Delta x^2)/(2*m*g)# #h=(42*(5/3)^2)/(2*0,28*9,81)# #h=(42*25/9)/(5,49)# #h=(116,67)/(5,49)# #h=21,25" m"#

Isabella Alvarez · Answer

undefined The maximum height ($h$) the ball will reach can be calculated using the formula:

$$ h = \frac{1}{2} \left( \frac{m}{k} \right) \left( \frac{v_0}{m} \right)^2 $$

where $m$ is the mass of the ball, $k$ is the spring constant, and $v_0$ is the initial velocity. The initial velocity ($v_0$) can be determined using the potential energy stored in the compressed spring:

$$ PE_{\text{spring}} = \frac{1}{2} k x^2 $$

where $x$ is the compression of the spring.

Given values:
- Mass ($m$): 280 g (convert to kg)
- Spring constant ($k$): 42 (kg/s^2)
- Compression of the spring ($x$): $\frac{5}{3}$ m

Calculate the initial velocity ($v_0$) using the potential energy formula and then use it to find the maximum height ($h$).