The launching mechanism of a toy gun consists of a spring ofunknown spring const

Question

The launching mechanism of a toy gun consists of a spring ofunknown spring constant, as shown in Figure P5.33a. If its springis compressed a distance of 0.200 mand the gun fired vertically as shown, the gun can launch a18.0 g projectile to a maximum heightof 16.0 m above the starting point ofthe projectile.

(a) Neglecting all resistive forces, determine thespring constant.
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(b) Neglecting all resistive forces, determine the speed of theprojectile as it moves through the equilibrium position of thespring (where x = 0), as shown in Figure 5.33b.
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Explanation / Answer

   According to law of conservation of energy,the kinetic energy of projectile at the time of launching will beexactly equal to its p.e. at highest point.    a.   P.E. at highestpoint   U   =   m* g * h   =   18.0 *10-3 * 9.8 *16.0   =   2.82   J          Alsoenergy stored in the spring   US   =   (1/2)* k * x2    k   =   springconstant,   x   =   compression       U   =   US   =   0.5* k * 0.2002       2.82   =   0.5* k * 0.04    =>   springconstant   k   =   2.82/0.02   =   141.0   N/m    b.   K.E.    =   P.E.       (1/2) * m *v2   =   2.82          0.5 *18.0 * 10-3 *v2   =   2.82          v   =   (2.82/ 9 *10-3)   =   17.70   m/s       2.82   =   0.5* k * 0.04    =>   springconstant   k   =   2.82/0.02   =   141.0   N/m    b.   K.E.    =   P.E.       (1/2) * m *v2   =   2.82          0.5 *18.0 * 10-3 *v2   =   2.82          v   =   (2.82/ 9 *10-3)   =   17.70   m/s

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