Question

A 50 g rock is launched vertically in the air by a slingshot. The slingshot exerts an average force of 53 N on the rock over a distance of 0.750 m. Find the speed of the rock as it leaves the slingshot in [m/s]? Find the maximum height reached by the rock in [m]. Neglect air resistance.

a) Speed: 14 m/s, Height: 1.02 m
b) Speed: 17 m/s, Height: 1.53 m
c) Speed: 20 m/s, Height: 2.04 m
d) Speed: 23 m/s, Height: 2.55 m

Answer 1

Calculating the work done by the slingshot and equating it to the kinetic energy provides the rock's launch speed, which is 39.87 m/s. The maximum height, found using the conservation of energy, is 80.62 m, which does not match any of the provided options.

Step-by-step explanation:

To find the speed of the rock as it leaves the slingshot, we first calculate the work done by the slingshot using work-energy principle:

Work = Force × Distance = 53 N × 0.750 m = 39.75 J

Work done by slingshot is converted entirely into the rock's kinetic energy since air resistance is neglected:

Kinetic Energy = ½ m v2

Substituting the work done for kinetic energy, we can solve for the speed (v):

39.75 J = ½ × 0.050 kg × v2 => v = √(39.75 J / 0.025 kg) = √(1590) ≈ 39.87 m/s

The maximum height reached by the rock can be found using the conservation of energy, where all kinetic energy at the start becomes potential energy at the highest point:

Potential Energy at max height = Kinetic Energy at beginning

mgh = ½ m v2

Assuming g = 9.8 m/s2, and rearranging for h:

h = v2 / (2g) = (39.87 m/s)2 / (2 × 9.8 m/s2) ≈ 80.62 m

From these calculations, it seems none of the options provided (a, b, c, d) in the question match the correct calculated values.