Question

A sprir1g with d orce constdnt o 5300 N/m and d rest ength go m)? (Assume the rock is launched from ground height. 2.9 m used ri dじd apuit when compressed to 1.0 m ' ı used launch d 50 kg rock. However there he release rnech n sm, 20 he rock ye s ldun hed d most stran h up. How hi h dues it an error in How fast is it going (in mys) when it hits tha ground? 84722x ms

Answer 1

(a) Using the law of conservation of energy,

initial potential energy of rock due to spring = gravitational potential energy of rock due to gravity

$\frac{1}{2}kx^2 = mgh$                    (1)

where k is spring constant, x is compression of spring, m is mass of rock, g is acceleration due to gravity, h is height of rock

compression of spring = rest length of spring - length of spring at compression

= 2.9 - 1.0 m

= 1.9 m

using values from problem and g = in (1),

9.8 m/s

53001.-58 *9.8h 5300 1.92 2 58 * 9.8 16.8 m

(b) velocity of ball when it hits the ground (v) is given using the equation,

since at maximum height, the ball gets stationary, u = 0 m/s

the ball falls due to gravity so a = g (acceleration due to gravity)

0+2* 9.8 16.8 u2329.3 v- V329.3 m/s - 18.1 m/s