Question

A 75 kg softball player does a head-first dive and is sliding to a stop to the

right across the infield dirt. Her initial velocity was 4 m/s, and she comes to
a stop in 8 seconds. Find all forces. Ignore air resistance.

Answer 1

The 75 kg softball player decelerates at -0.5 m/s² due to the frictional force while sliding to a stop, resulting in a force of -37.5 N acting on the player.

Step-by-step explanation:

The student has described a scenario where a 75 kg softball player slides to a stop across the infield dirt with an initial velocity of 4 m/s over 8 seconds. Since air resistance is ignored, we'll consider only the force due to friction that brings the player to a stop. Using Newton's second law, F = ma, where F is the force, m is the mass, and a is the acceleration, we can calculate the deceleration and then the force.

First, we find the acceleration (which is actually deceleration since the player is coming to a stop):
a = (vf - vi) / t
Where vf is the final velocity (0 m/s), vi is the initial velocity (4 m/s) and t is time (8 s). Thus, acceleration a = (0 - 4) m/s2 / 8 s = -0.5 m/s2.

To find the force of friction, we use:
F = m * a
Here, m = 75 kg and a = -0.5 m/s2, so the force is F = 75 kg * (-0.5 m/s2) = -37.5 N. The negative sign indicates that the force is opposite in direction to the player's initial motion.

Answer 2

vf = vi + at

vf = 0

a = - (to stop (deceleration))

input the value:

0 = 4 m/s -a.8

8a = 4

a = 0.5 m/s²

F = m.a

F = 75 kg x 0.5 m/s²

F = 37.5 N