4. A roller coaster starts at the top of a hill, 42 m above ground, with a velocity of 5.0 m/s.
a. What is the velocity of the roller coaster of mass 830 kg, further down the track where it is 16 m
above ground? *neglecting friction [3]
b. The roller coaster having a velocity from part a, now travels over level track with a coefficient of
friction of 0.87 and eventually comes to rest. How long does the track need to be? Explain your
solution. [4]
c. Explain the energy changes throughout the rollercoaster’s trip, from the top of the hill to when it
stops. [3]

A roller coaster starts from rest at a point 45 m above the bottom of a dip (See Fig. 6-2). Neglect friction, what will be the speed of the roller coaster at the top of the next slope, which is 30 m above the bottom of the dip?

a. 17 m/s b. 24 m/s c. 14 m/s d. 30 m/s

You have no mass given, so you work it cancelling the mass:

i'll take g = 9.8 m/s^2

(g x h1) - (g x h2) = 1/2v^2 ( no mass either for KE)

(9.8 x 45) - (9.8 x 30) = 147

147 x 2 = v^2 = 294, sq-rt = v = 17.146 m/s answer (17m/s)

A roller coaster starts with a speed of 5.0 m/s at a point 45 m above the bottom of a dip (See Fig. 6-2). Neglect friction, what will be the speed of the roller coaster at the top of the next slope, which is 30 m above the bottom of the dip? a. 16 m/s b. 12 m/s c. 14 m/s d. 18 m/s

u^2 = 2gh (5^2 = 2gh)

25/(2g) = h = 1.2755 m

45 + 1.2755 = 46.2755m

Do the same equations as question one, answer = 17.86 m/s (answer 18 m/s)

A roller coaster starts at a point 30 m above the bottom of a dip with a speed of 25 m/s (See Fig. 6-2). Neglect friction, what will be the speed of the roller coaster at the top of the next slope, which is 45 m above the bottom of the dip? a. 14 m/s b. 16 m/s c. 20 m/s d. 18 m/s

Do the same! 25^2 = 2gh, 625/(2g) = h = 31.89 m

31.89 + 30 = 61.89 m

(9.8 x 61.89) - (9.8 x 45) = 1/2v^2 = 165.52

165.52 x 2 = 331.04, sq-rt = 18.19 m/s (answer 18 m/s)

A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final speed of the cart?

a. 73 m/s b. 6.0 m/s c. 36 m/s d. 8.5 m/s

Ignoring friction:

F = ma

200/55 = a = 3.636 m/s^2

v^2 = u^2 + 2as, u^2 = zero

v^2 = 2as

2as = 2 x 3.636 x 10 = 72.72, sq-rt = 8.52 m/s (answer 8.5 m/s)

Explanation: