A 73-kg base runner begins his slide into second base when he is moving at a spe

Question

A 73-kg base runner begins his slide into second base when he is moving at a speen of 3.8 m/s. The coeffieient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as the reaches the base. (a) How much mechanical energy is list due to friction action on the runner? j (b) How far does he slide? m A 0.21-kg stone is held 1.1 m above the top edge of a water well and then dropped into it. The well has a depth of 5.1 m. (a) Taking y - 0 at the top edge of the well, what is the gravitational potential energy of the stone-Earth system before the stone is released? j (b) Taking y = 0 at the top edge of the well, what is the gravitational potential energy of the stone-Earth system when it reaches the bottom of the well? j What is the change in gravitational potential energy of the system from release to reaching the bottom of the well? j A daredevil on a motorcycle leaves the end of a ramp with a speed of 34.3 m/s as in the figure below. If his speed is 32.4 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance. m

Explanation / Answer

7)

Given that, mass of the man is m=73 kg

velocity of the runner is,v=3.8 m/s2

a)

The mechanical energy is,

E= 1/2 mv^2

substitute the values in above,

so, E = 1/2 (73 kg) (3.8 m/s2)^2
then, we get 527,

because he is slowing down, it will be -527 J

b)

The Force of friction is,

F=uk* N

where N is the normal force=mg
F= 0.70 * 73 kg * 9.8 m/s2
we get F=500.78 N

then,W=Fd
so rearrange the term,

d=W/F=(527 J)/(500.78 N)
d= 1.05 m

8)

Given that,

mass of the stone is, m=0.21 kg

height,h=1.1 m

depth,d=5.1 m

a)

The potential energy is,

P.E1=m*g*h

     =0.21 kg*9.8 m/s2*1.1 m
     = 2.26 J

b)

The potential energy is,

P.E 2=m*g*d

      =0.21 kg *9.8 m/s2*(-5.1 m)
      = -10.49 J

c)

Change is,P.E = P.E2 - P.E1
                      = -10.49 J - 2.26 J
                      =-12.75 J

9)

Given that,

v1=34.3 m/s and v2=32.4 m/s

from the conservation of energy,

1/2mv12+mgh1=1/2mv22+mgh2

but mass is cancel , initial height is ,h1=0 then rearrange the terms,

h2=(v12-v22)/g

substitute the values in above,

h2=((34.3 m/s)2-(32.4 m/s)2)/(9.8 m/s2)

    =12.93 m

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