The launching mechanism of a toy gun consists of a spring of unknown spring cons

Question

The launching mechanism of a toy gun consists of a spring of unknown spring constant, as shown in Figure P5.33a. If its spring is compressed a distance of 0.160 m and the gun fired vertically as shown, the gun can launch a 20.0 g projectile to a maximum height of 20.0 m above the starting point of the projectile.


(a) Neglecting all resistive forces, determine the spring constant.

(b) Neglecting all resistive forces, determine the speed of the projectile as it moves through the equilibrium position of the spring (where x = 0), as shown in Figure 5.33b.

Explanation / Answer

a) E.P.E=G.P.E
E.P.E=1/2XforceXextension
Force = Kx
E.P.E= 1/2X(k)X(Extension)^2

G.P.E=MXgrav acc. X height
G.P.E=(20/1000)(9.8)(20.0)=3.92 J
3.92 = 1/2(K)(0.120)^2
K(spring constant ) =544.4444444 Nm^-1

b) V^2=2as+U^2
s=20.0
a=9.8
u = 0 (because at max height velocity is 0)
V^2=2(9.8)(2)
V^2=392
V=19.79898987 ms^-1

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