How much potential energy was stored in the compressed spring

Asked by admin @ 22/11/2021 in Physics viewed by 294 People

The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how much is the spring is compressed. Submit Answer Tries 0/20 A 0.050 kg dart is fired straight up. Find the vertical distance the dart travels from its position when the spring is compressed to its highest position. Submit Answer Tries 0/20 The same dart is now fired horizontally from a height of 3.90 m. The dart remains in contact until the spring reaches its equilibrium position. Find the horizontal velocity of the dart at that time. Submit Answer Tries 0/20 Find the horizontal distance from the equilibrium position at which the dart hits the ground.

Answered by admin @ 22/11/2021

Answer:

A

$x = 0.198456 \ m$

B

$h = 1.3061 \ m$

C

$v = 5.06 \ m/s$

D

$d = 4.0273 \ m$

Explanation:

Considering the first question

From the question we are told that

The spring constant is $k = 32.50 N/m$

The potential energy is $PE = 0.640 \ J$

Generally the potential energy stored in spring is mathematically represented as $PE = \frac{1}{2} * k * x^2$

=> $0.640= \frac{1}{2} * 32.50 * x^2$

=> $x = \sqrt{0.03938}$

=> $x = 0.198456 \ m$

Considering the second question

From the question we are told that

The mass of the dart is m = 0.050 kg

Generally from the law of energy conservation

$PE = mgh$

=> $0.640 = 0.050 * 9.8 * h$

=> $h = 1.3061 \ m$

Considering the third question

The height at which the dart was fired horizontally is $H = 3.90\ m$

Generally from the law of energy conservation

$PE = KE$

Here KE is kinetic energy of the dart which is mathematical represented as

$KE = \frac{1}{2} * mv^2$

=> $0.640 = \frac{1}{2} * 0.050 * v^2$

=> $v^2 = 25.6$

=> $v = 5.06 \ m/s$

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

$t = \frac{ 2 * H }{g}$

=> $t = \frac{ 2 * 3.90 }{9.8 }$

=> $t = 0.7959 \ s$

Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as

$d = v * t$

=> $d = 5.06 * 0.7959$

=> $d = 4.0273 \ m$

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