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3 years ago

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A 75 kg baseball player runs at a velocity of 6 m/s before sliding to a stop at second base. a. What is the kinetic energy of th

e runner before he begins his slide? b. What is the kinetic energy of the runner once he reaches the base? c. What is the change in the kinetic energy of the runner? d. How much work is done by friction in stopping the runner? e. If the runner slides for 2 m, what is the force of friction that acts upon him?

1 answer:

lana [24]3 years ago

6 0

Answer:

a. $\displaystyle k_o=1350\ J$

b. $\displaystyle k_1=0\ J$

c. $\Delta k=-1350\ J$

d. $W=-1350\ J$

e. $F=-675\ N$

Explanation:

<u>Work and Kinetic Energy </u>

When an object moves at a certain velocity v0 and changes it to v1, a change in its kinetic energy is achieved:

$\Delta k=k_1-k_0$

Knowing that

$\displaystyle k=\frac{mv^2}{2}$

We have

$\displaystyle \Delta k=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

The work done by the force who caused the change of velocity (acceleration) is

$\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

If we know the distance x traveled by the object, the work can also be calculated by

$W=F.x$

Being F the force responsible for the change of velocity

The 75 kg baseball player has an initial velocity of 6 m/s, then he slides and stops

a. Before the slide, his initial kinetic energy is

$\displaystyle k_o=\frac{mv_0^2}{2}$

$\displaystyle k_o=\frac{(75)6^2}{2}$

$\boxed{\displaystyle k_o=1350\ J}$

b. Once he reaches the base, the player is at rest, thus his final kinetic energy is

$\displaystyle k_1=\frac{(75)0^2}{2}$

$\boxed{\displaystyle k_1=0\ J}$

c. The change of kinetic energy is

$\Delta k=k_1-k_0=0\ J-1350\ J$

$\boxed{\Delta k=-1350\ J}$

d. The work done by friction to stop the player is

$W=\Delta k=k_1-k_0$

$\boxed{W=-1350\ J}$

e. We compute the force of friction by using

$W=F.x$

and solving for x

$\displaystyle F=\frac{W}{x}$

$\displaystyle F=\frac{-1350\ J}{2\ m}$

$\boxed{F=-675\ N}$

The negative sign indicates the force is against movement

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