186k views
0 votes
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: 106 m/s, Height: 112.5 m.
B. Speed: 80 m/s, Height: 75 m.
C. Speed: 106 m/s, Height: 75 m.
D. Speed: 80 m/s, Height: 112.5 m.

1 Answer

7 votes

Final answer:

The speed of the rock as it leaves the slingshot is approximately 106 m/s. The maximum height reached by the rock is 2.20 m.

Step-by-step explanation:

To find the speed of the rock as it leaves the slingshot, we can use the work-energy theorem. The work done on the rock by the slingshot is equal to the change in its kinetic energy. The work done is the product of the force and the distance over which it acts. So, the force exerted by the slingshot is equal to the weight of the rock. The speed of the rock as it leaves the slingshot is given by the equation:

Speed = √(2 * force * distance / mass)

Substituting the given values:

Speed = √(2 * 53 N * 0.750 m / 0.050 kg) ≈ 106 m/s

To find the maximum height reached by the rock, we can use the principle of conservation of mechanical energy. At the maximum height, the kinetic energy of the rock is zero and all its initial potential energy is converted to gravitational potential energy. The gravitational potential energy is given by the equation:

Potential energy = mass * acceleration due to gravity * height

At the maximum height, the potential energy is equal to the initial potential energy:

0.050 kg * 9.8 m/s^2 * height = 0.050 kg * 9.8 m/s^2 * 2.20 m

Simplifying, we find:

Height = 2.20 m

User Thomas Chilinski
by
7.9k points