Question

A 92.0kg baseball player slides into second base. The coefficient of friction between the player and the ground is 0.61. If the player comes to rest after 1.2s, what was his initial speed?

Answer 1

Answer:

The initial speed is 7.174 m/s.

Explanation:

Given that,

Mass = 92.0 kg

Coefficient of friction = 0.61

Time = 1.2 s

We need to calculate the acceleration

Using of friction force

$F = \mu mg$...(I)

Where, $\mu$ =Coefficient of friction

g = acceleration due to gravity

Using newton's law

$F = ma$....(II)

m = mass of the baseball

a = acceleration of the baseball

From equation (I) and (II)

$ma=\mu mg$...(I)

Put the value in the equation

$a=0.61\times9.8$

$a=5.978\ m/s^2$

We need to calculate the initial velocity

Using equation of motion

$v = u-at$

Where, v = final velocity

u = initial velocity

a = acceleration

Here, a is negative because the player comes to rest

t = time

Put the value in the equation

$0=u-5.978\times1.2$

$u=7.174\ m/s$

Hence, The initial speed is 7.174 m/s.