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A bowling ball of mass m = 1.1 kg is resting on a spring compressed by a distance d = 0.35 m when the spring is released. At the moment the spring reaches its equilibrium point, the ball is launched from the spring into the air in projectile motion at an angle of θ = 39° measured from the horizontal. It is observed that the ball reaches a maximum height of h = 4.4 m, measured from the initial position of the ball. Let the gravitational potential energy be zero at the initial height of the bowling ball.
a) what is the spring constant, k, in newtons per meter?

(I got that the speed of the ball after the launch is 14.76)

Respuesta :

onyebuchinnaji

The spring constant, k, in newtons per meter is 1,955.9 N/m.

Speed of the ball after the launch

h = v²sin²θ/2g

v = √[(2gh)/sin²θ]

v = √[(2 x 9.8 x 4.4)/ (sin 39)²]

v = 14.76 m/s

Energy of the ball at top

E = K.E + P.E

E = ¹/₂m(v cosθ)²  +  mgh

E =  ¹/₂(1.1)(14.76 cos39)²  +  (1.1 x 9.8 x 4.4)

E = 119.8 J

Spring constant

E = ¹/₂kx²

k = 2E/x²

k = (2 x 119.8)/(0.35²)

k = 1,955.9 N/m

Thus, the spring constant, k, in newtons per meter is 1,955.9 N/m.

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