A slingshot is used to launch a 5.00-gram rock.The effective spring constant of the slingshot is 200 N/m.By what distance must the slingshot be stretched in order to launch the rock at 20.0 m/s?

Question

Quizplus · Accepted Answer

The potential energy stored in the slingshot is given by U = (1/2)kx^2, where k is the spring constant and x is the distance stretched. We can rearrange this equation to solve for x:
x = sqrt(2U/k)
To launch the rock at 20.0 m/s, the initial kinetic energy of the rock must be (1/2)mv^2 = 0.5 * 0.005 kg * (20.0 m/s)^2 = 2 J. This is also the amount of potential energy that must be stored in the slingshot before launch. Therefore:
x = sqrt(2 * 2 J / 200 N/m) = 0.1 m = 10 cm
Therefore, the slingshot must be stretched by 10 cm (or 0.1 m) to launch the rock at 20.0 m/s. The correct answer is C.

A Slingshot Is Used to Launch a 5