Question

A 0.58-kg object travels from point A to point B. If the speed of the object at point A is 6.0 m/s and the kinetic energy at point B is 8.0 J, determine the following.

(a) kinetic energy of the object at point A
(b) speed of the object at point B

Answer 1

The kinetic energy of the 0.58-kg object at point A is 10.44 J, and the speed of the object at point B is 5.25 m/s.

Step-by-step explanation:

To determine (a) the kinetic energy of the object at point A and (b) the speed of the object at point B when A 0.58-kg object travels from point A to point B, we can use the formula for kinetic energy (KE = 0.5 * m * v2).

(a) Kinetic energy at point A:

• KEA = 0.5 * m * vA2
• KEA = 0.5 * 0.58 kg * (6.0 m/s)2
• KEA = 0.5 * 0.58 kg * 36 m2/s2
• KEA = 10.44 J

(b) Speed of the object at point B:

• KEB = 8.0 J
• 8.0 J = 0.5 * 0.58 kg * vB2
• vB2 = (8.0 J) / (0.5 * 0.58 kg)
• vB2 = 27.59 m2/s2
• vB = √27.59 m2/s2
• vB = 5.25 m/s