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.

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