If it take a force of $95 N$ to compress the spring of a toy Pop gun $0.175 m$ to load a $0.160 k g$ ball,with what speed will the ball leave the gun if fired horizontally?

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Solution

Given, Spring Force $= 95 N = k x$ {where k is the spring constant, $x = 0.175 m$ }

$\frac{95}{0.175} = k \Rightarrow 542.85 N / m = k$
Spring P.E at $0.175 m$ compression $= \frac{1}{2} k x^{2} = (0.5) (542.85) (0.175)^{2} = 8.3123 J$
KE of ball $= \frac{1}{2} m V^{2}$
$\Rightarrow 8.3123 = (0.5) (0.16) V^{2}$
$\Rightarrow V^{2} = \frac{8.3123}{(0.5) (0.160)} = 103. 903$
$\Rightarrow V = 10.1933 m / s$

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If it take a force of 95N to compress the spring of a toy Pop gun 0.175m to load a 0.160kg ball,with what speed will the ball leave the gun if fired horizontally?

Given, Spring Force =95N=kx {where k is the spring constant, x=0.175m} 950.175=k⇒542.85N/m=k Spring P.E at 0.175m compression =12kx2=(0.5)(542.85)(0.175)2=8.3123JKE of ball =12mV2 ⇒8.3123=(0.5)(0.16)V2 ⇒V2=8.3123(0.5)(0.160)=103.903 ⇒V=10.1933m/s

If it take a force of $95 N$ to compress the spring of a toy Pop gun $0.175 m$ to load a $0.160 k g$ ball,with what speed will the ball leave the gun if fired horizontally?