Please show every step of the solutions. Many thanks in advance! A baseball outf

Question

Please show every step of the solutions. Many thanks in advance!

A baseball outfielder throws a 0.15 kg baseball at a speed of 40 m/s and an initial angle of 30degree. (a) What is the kinetic energy of the baseball at the highest point of its trajectory? (b) What is the kinetic energy just before it hits the ground? (c) How much work is done by the gravitational force during its time in the air? You may ignore air resistance. Let's assume that a rock is sliding across the "Racetrack Playa" in Death Valley National Park and that it has an initial speed v. As a result of kinetic friction, it comes to rest after traveling a distance d. Determine the stopping distance of the rock if it had started with an initial speed of 3v instead. You should be solving/explaining the answer to this using the Work Energy Theorem. A box has an initial speed of 10 m/s. It slides across a flat section and then up an incline. Use L = 20 m, mu k = 0.1 and theta = 40degree. Note that I have not given you the mass. It will cancel out of all the equations in the end! (a) Determine the speed of the box at the end of the flat section, (b) Determine the distance D it slides up the incline before coming to a complete stop. A 5 kg block is pushed 3 m up a wall by a 90 N force F. Assume that the block starts from rest and that it has an acceleration of 0.5 m/s2. Use theta = 55degree. (a) Determine the coefficient of kinetic friction between the block and the wall, (b) Find the work done by each force acting on the block, (c) Use the work energy theorem to determine the speed of the block after it has moved those 3 m.

Explanation / Answer

1) a)

at max height vy = 0

vx = 40*cos(30 degrees)

KE = 0.5*0.15*(40*cos(30 degrees))^2= 90 J

b) KE = 1/2 mv^2 = 0.5*0.15*40^2= 120 J

c) since conservative, W = 0

2) W = dE

- F d = 1/2 mv^2

so if 3 v

- F x = 1/2 m (3 v)^2 = 9 1/2 mv^2 = 9* - F d

x = 9 d

3)

a) Ei + W friction = Ef

1/2 mv^2 - u m g d = 1/2 mv^2

0.5*10^2 - 0.1*9.81*20 = 0.5*v^2

v=7.79 m/s

b)

Ei + W = Ef

1/2 mv^2 - u m g cos theta d = m g h

0.5*7.79^2 - 0.1*9.81*cos(40 degrees)*d = 9.81*d*sin(40 degrees)

d=4.3 m

5)

sum forces in the x direction

P cos theta - N = 0

N = P cos theta

sum in the y

P sin theta - mg - friction = m a

P sin theta - m g - u P cos theta = m a

90*sin(55 degrees) - 5*9.81 - u*90*cos(55 degrees) = 5*0.5

u=0.43

b) W gravity = - m g d = -5*9.81*3= - 147 J
W Force = P d sin theta = 90*3*sin(55 degrees)=221 J

W friction = - friction d = -0.43*90*cos(55 degrees)*3= - 67 J

c) Wnet = -147*221-67= 7 J

so KE = Wnet

0.5*5*v^2 = 7

v=1.73 m/s

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