Sping Constant?

Question

In a physics lab experiment, a spring clamped to the table is used to shoot a 45g ball at a 33 degrees angle. When the spring is compressed 17 cm, the call travels horizontally 6.0 m and lands 1.70 m below the point at which it left the spring. What is the spring constant?

Answer 1

1. Energy of the spring is converted to kinetic energy of the ball
Pe=Ke
Pe= 0.5k x^2
Ke=0.5 m V^2
k= (V/x)^2
Now we have to find V

V is the vector product of horizontal and vertical components Vh and Vv respectively
V=sqrt(Vh^2 + Vv^2) since at the instant the ball was fired Vv=Vh tan(33)
V=Vh sqrt(1+tan^2(33))

or
Vh=Vcos(33) and
Vv=Vsin(33)

Now lets write some more equation describing the motion
Vh=S/t where
S - 6m
t - total time in flight


h= Vvt – 0.5 g t^2

t= [Vv + sqrt(Vv - 2 hg)]/g
and since t= S/Vh
S/Vh=[Vv + sqrt(Vv - 2 hg)]/g
S=Vh[Vv + sqrt(Vv - 2 hg)]/g

Now substitute Vh and Vv
S=Vcos(33)[Vsin(33) + sqrt(Vsin(33) -2 hg)]/g
Since
S=6.0m
h= -1.70m
g=9.8m/s^2

We have to solve for V and substitute its value into k= (V/x)^2


Have fun