Question

1) A 67-kg base runner begins his slide into second base when he is moving at a speed of 3.7 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base. (a) How much mechanical energy is lost due to friction acting on the runner? (b) How far does he slide?

2) A running 53-kg cheetah has a top speed of 31 m/s. (a) What is the cheetah's maximum kinetic energy? (b) Find the cheetah's speed when its kinetic energy is one half of the value found in part (a).

3) A 7.80-g bullet moving at 500 m/s penetrates a tree trunk to a depth of 5.10 cm. (a) Use work and energy considerations to find the average frictional force that stops the bullet. (b) Assuming the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment it stops moving.

4) A daredevil on a motorcycle leaves the end of a ramp with a speed of 31.5 m/s as in the figure below. If his speed is 29.5 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance.

Answer 1

1a) lost in mechanical energy = -0.5*m*u^2 = -0.5*67*3.7*3.7 = -458.615 J

b) work done by friction = 458.615

mu_k*m*g*S = 458.615

S= 458.615/(0.7*67*9.81) = 0.996 m = 1m

2) A) KEmax = 0.5*m*v^2 = 0.5*53*31*31 = 25466.5 J

b) if KE = KEmax/2 =25466.5/2 = 0.5*m*u^2 = 0.5*53*u^2

u = 21.9 m/s = 22 m/s

3) f*S = 0.5*m*v^2 = 0.5(7.8*10^-3)*500*500 = 975 J

frictional force is f = 975/S = 975/0.051 = 19117.64 N

4) Horizontal component of original velocity = 31.5 cosø
where ø is the angle of takeoff.

31.5 cosø = 29.5

ø = 20.52˚

Vertical component of velocity = 31.5 * sin (20.52) = 11.0 m/s

Height reached = V^2 / 2 . g = 11^2 / (2 * 9.8) = 6.13 m