Question

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.

Answer 1

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

       $d = v * t$

=>     $d = 5.06 * 0.7959$

=>     $d = 4.0273 \ m$