Abowling ball of mass m=1.7kg is launched from a spring compressed by a distance d=0.31m at an angle of theta=37 measured from the horizontal. it is observed that the ball reaches a maximum height of h=4.9m, measured from the initial position of the ball. let the gravitational potential energy be zero at the initial height of the bowling ball.

part a) what is the spring constant k, in newtons per meter?

part b) calculate the speed of the ball, v0 in m/s, just after the launch.

k = 1 700.7 N/m

v0 = 9.8 m/s^2

Explanation:

Hello!

We can answer this question using conservation of energy.

The potential energy of the spring (PS) will transform to kinetic energy (KE) of the ball, and eventually, when the velocity of the ball is zero, all that energy will be potential gravitational (PG) energy.

When the kinetic energy of the ball is zero, that is, when it has reached its maximum heigh, all the potential energy of the spring will be equal to the potential energy of the gravitational field.

PS = (1/2) k x^2  where x is the compresion or elongation of the spring

PG = mgh

a)

Since energy must be conserved and we are neglecting any energy loss:

PS = PG

Solving for k

k = (2mgh)/(x^2) = ( 2 * 1.7 * 9.81 * 4.9 Nm)/(0.31^2 m^2)

k = 1 700.7 N/m

b)

Since the potential energy of the spring transfors to kinetic energy of the ball we have that:

PS = KE

that is:

(1/2) k x^2 = (1/2) m v0^2

Solving for v0

v0 = x √(k/m) = (0.31 m ) √( 1 700.7 N/m / 1.7kg)

v0 = 9.8 m/s^2

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer