Question

A 12.0-g rock is placed in a slingshot with a spring constant of 200 N/m. The rock is stretched back 0.500 m. After the slingshot is fired, what is the maximum velocity of the rock?
25.0 m/s

64.5 m/s

0.600 m/s

82.1 m/s

Answer 1

The maximum velocity of the rock is 64.55 m/s

Answer: Option B

Explanation:

The possessed energy by the objects while motion called kinetic energy. It is directly proportionate to mass and square of velocity. The stored energy in the system called potential energy and expressed as below,

$\text { potential energy }=\frac{1}{2} \times(\text { spring constant }) \times\left(\text { distance from equilibrium) }^{2}\right.$

                  $U=\frac{1}{2} \times k \times x^{2}$

U = spring’s potential energy in certain place

k = the spring constant, in N/m.

x = the spring’s distance is stretched or compressed away from equilibrium

Conservation of energy states energy neither created nor destroyed but change from one form to other. So, using this concept, equating both potential and kinetic energies equation we get,

            $\frac{1}{2} \times m \times v^{2}=\frac{1}{2} \times k \times x^{2}$

Given:

m = 12 g = 0.012 kg

k = 200 N/m

x = 0.500 m

Substitute these values we get,

         $\frac{1}{2} \times 0.012 \times v^{2}=\frac{1}{2} \times 200 \times(0.500)^{2}$

         $v^{2}=\frac{200 \times 0.25}{0.012}=\frac{50}{0.012}=4166.66$

Taking square root, we get,

         $v=64.55 \mathrm{m} / \mathrm{s}$