Question

A particle of mass m = 5 kg is released from point A on a frictionless track at height ha = 5.6 m. The acceleration of gravity is 9.8 m/s². Determine the particle's speed at point B at height hb = 1.7 m. Answer in units of m/s.

A) 13.1 m/s
B) 9.2 m/s
C) 3.5 m/s
D) 7.7 m/s

Answer 1

The particle's speed at point B is 13.1 m/s.

Step-by-step explanation:

To determine the particle's speed at point B, we can use the principle of conservation of mechanical energy. Since the track is frictionless, the mechanical energy of the particle is conserved as it moves from point A to point B. When the particle is at point A, it has gravitational potential energy which will be converted to kinetic energy at point B.

Using the equation for gravitational potential energy, mgh, where m is the mass, g is the acceleration due to gravity, and h is the height, we can calculate the initial potential energy at point A: (5 kg)(9.8 m/s²)(5.6 m).

At point B, the potential energy will be converted to kinetic energy, given by the equation (1/2)mv², where m is the mass and v is the velocity or speed. Equating the initial potential energy to the final kinetic energy, we can solve for the speed at point B.

1. Initial potential energy at point A: (5 kg)(9.8 m/s²)(5.6 m)
2. Final kinetic energy at point B: (1/2)(5 kg)v²
3. Equating the two energies:
4. (5 kg)(9.8 m/s²)(5.6 m) = (1/2)(5 kg)v²
5. Simplifying and solving for v, we get v = 13.1 m/s.