Question

A 70.0 kg base runner moving at a speed of 4.0 m/s begins his slide into second base. The coefficient of friction between his clothes and Earth is 0.70. His slide lowers his speed to zero just as he reaches the base. How much mechanical energy is lost because of friction acting on the runner? A. 1100 J B. 560 J C. 140 J D. 0J 8. How far does the runner slide? F. 0.29 m G. 0.57 m H. 0.86 m J. 1.2 m

Answer 1

Answer:

B. 560 J

J. 1.2 m

Explanation:

v = Final velocity = 0

u = Initial velocity = 4 m/s

$\mu$ = Coefficient of friction = 0.7

m = Mass of runner = 70 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Kinetic energy is given by

$K=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow K=\dfrac{1}{2}\times 70\times (0^2-4^2)\\\Rightarrow K=-560\ \text{J}$

The mechanical energy lost is 560 J

Acceleration is given by

$a=-\mu g\\\Rightarrow a=-0.7\times 9.81\\\Rightarrow a=-6.867\ \text{m/s}^2$

From kinematic equations we get

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-4^2}{2\times -6.867}\\\Rightarrow s=1.165\approx 1.2\ \text{m}$

The runner slides for 1.2 m