Question

After hitting a long fly ball that goes over the right fielder's head and lands in the outfield, the batter decides to keep going past second base and try for third base. The 60 kg player begins sliding 3.40 m from the base with a speed of 4.35 m/s.
a) How much work was done on the player by friction?
b) What was the coefficient of kinetic friction between the player and the ground?

Answer 1

Answer:

A) -567.675 J

B) 0.284

Explanation:

Parameters given:

Mass of player, m = 60 kg

Distance slid, d = 3.40 m

Speed = 4.35 m/s

A) We know that work done is equal to change in kinetic energy of a body:

W = ΔKE

Change in kinetic energy is given as:

$\Delta KE =0 - 0.5mv^2\\\\ \Delta KE = -0.5mv^2$

where m = mass

v = velocity

=>W =  ΔKE = -0.5 * 60 * 4.35 * 4.35 = -567.675 J

The work is negative because the force of friction acts opposite the motion of the player.

B) We have that:

Fr = μmg

where Fr = Frictional force

μ = coefficient of kinetic friction

g = acceleration due to gravity

Work done is give as:

W = Fr * d

(d = distance moved)

=> Fr = W/d

Therefore:

W / d = μmg

=>μ = W / (m * g * d)

μ = (567.675) / (60 * 9.8 * 3.40)

μ = 0.284

Answer 2

The work done on the player by the friction is -567.67 J and coefficient of kinetic friction  is 0.248.

Given here,

Mass of player= 60 kg  

Slide Distance  = 3.40 m  

Speed = 4.35 m/s

(A) The work done by friction is equal to the kinetic energy of the system or body,

$\bold {W = \Delta KE}$

$\bold {\Delta KE = 0 - 0.5 mv^2}\\\\\bold {\Delta KE = - 0.5 mv^2}$

So, work done by friction  on the player,

$\bold { -0.5 \times 60 \times (4.35)^2 = -567.675 J}$

negative sign shows that the force of friction acts opposite the motion of the player.

(B) The friction can be calculated by the formula,

$\bold {F_f = \mu mg}$

Where,

$\bold {F_f}$ - Frictional force  

$\bold {\mu}$ = coefficient of kinetic friction  

g = gravitational acceleration

Work done by friction can be calculate as,

$\bold {W = F_f \times d}$

$\bold {F_f =\dfrac {W}{d}}$

So coefficient of kinetic friction,   

$\bold{\mu = \dfrac {W}{mgd}}$

Put the value and calculate the formula,

$\bold {\mu = \dfrac {567.675}{60 \times 9.8 \times 3.40}}\\\\\bold {\mu = 0.248}$

Therefore, work done on the player by the friction is -567.67 J and coefficient of kinetic friction  is 0.248.

To know more about friction,

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