Question

A 0.157 kg ball is thrown straight up from 2.33 m above the ground. Its initial vertical speed is 11.60 m/s. A short time later, it hits the ground. Calculate the total work done by the force of gravity during that time.

Answer 1

Answer:

14.148 J

Explanation:

Work : This can be defined as the product of force and the distance moved along the direction of force. The S.I unit of work is Joules.(J).

From the question,

The Total work done by the field = Kinetic energy of the ball + Potential energy of the ball

Wt = 1/2mv²+mgh......................... Equation 1

Where Wt = Total work done by the ball, m = mass of the ball, v = velocity of the ball, g = acceleration due to gravity of the ball, h = height of the ball.

Given: m = 0.157 kg, v = 11.60 m/s, h = 2.33 m, g = 9.8 m/s²

Substitute these values into equation 1

Wt = 1/2(0.157)(11.6²) + (0.157)(2.33)(9.8)

Wt = 10.563+3.585

Wt = 14.148 J