1) A 0.200-kg ball is loaded into a popgun.The ball is loaded by compressing a s

Question

1) A 0.200-kg ball is loaded into a popgun.The ball is loaded by compressing a spring 0.185 m. The spring isinitially not compressed, then, as it is compressed, the forceincreases up to a maximum of 161 N. The gun is then fired. What isthe speed (in m/s) of the ball as it leaves the gun? (Ans:1.22 × 101) 2) A box with mass 0.9 kg is attached to aspring with spring constant k = 730 N/m. The box oscillates insimple harmonic motion. At the moment when the box is moving with aspeed of 0.63 m/s, it is 0.03 m from its equilibrium position. Whatis the amplitude (in m) of the motion? (Answer: 3.73x10^ -2) 1) A 0.200-kg ball is loaded into a popgun.The ball is loaded by compressing a spring 0.185 m. The spring isinitially not compressed, then, as it is compressed, the forceincreases up to a maximum of 161 N. The gun is then fired. What isthe speed (in m/s) of the ball as it leaves the gun? (Ans:1.22 × 101) 2) A box with mass 0.9 kg is attached to aspring with spring constant k = 730 N/m. The box oscillates insimple harmonic motion. At the moment when the box is moving with aspeed of 0.63 m/s, it is 0.03 m from its equilibrium position. Whatis the amplitude (in m) of the motion? (Answer: 3.73x10^ -2) 2) A box with mass 0.9 kg is attached to aspring with spring constant k = 730 N/m. The box oscillates insimple harmonic motion. At the moment when the box is moving with aspeed of 0.63 m/s, it is 0.03 m from its equilibrium position. Whatis the amplitude (in m) of the motion? (Answer: 3.73x10^ -2)

Explanation / Answer

The spring constant is              k= (161 N)/(0.185 m)                 = 870.27 N/m From the law of conservation of energy            1/2 kx2 = 1/2 mv2     The speed of the ball is                   v = (kx2/m) Here x = 0.185 m           m=0.200 kg               v = 1.22 × 101 m/s 2)    The speed of the box is                v =[A2 -x2]              (0.63 m/s) =((730 N/m)/(0.90kg))[A2 -(0.03m )2]                    A = 3.73x10^ -2 m               The speed of the ball is                   v = (kx2/m) Here x = 0.185 m           m=0.200 kg               v = 1.22 × 101 m/s 2)    The speed of the box is                v =[A2 -x2]              (0.63 m/s) =((730 N/m)/(0.90kg))[A2 -(0.03m )2]                    A = 3.73x10^ -2 m          

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