A 1000. kg roller coaster travels over a 20.0 m tall hill at 2.00 m/s. a) Later

Question

A 1000. kg roller coaster travels over a 20.0 m tall hill at 2.00 m/s. a) Later it coasts over a 15.0 m tall hill. What is its highest possible speed as it crests that hill? b) Later it runs through a 10.0 m long flat section of track where mu_k = 0.300. How much mechanical energy is lost? c) If it later coasts to a stop on a third hill, how high could that hill be, if the only mechanical energy loss was in part b)? A student driving nails into a piece of wood lifts a 2.25 kg hammer 0.300 m above the and then simply drops it on the nail. If the nail is driven 0.0100 m into the wood, calculate the average force on the hammer.

Explanation / Answer

Here ,

m = 1000 Kg

hi = 20 m

u = 2 m/s

a) let the speed is v

if there are no loses

using conservation of energy

initital kinetic energy + intiail potential energy = final kinetic energy + final potential energy

0.50 * 1000 * v^2 + 1000 * 9.8 * (15) = 0.50 * 1000 * 2^2 + 1000 * 9.8 * 20

solving for v

v = 10.1 m/s

the speed of roller coaster is 10.1 m/s

b)

mechanical energy lost = work done by friction

mechanical energy lost = frictional force * distance

mechanical energy lost = u * m * g * distance

mechanical energy lost = 0.30 * 1000 * 9.8 * 10

mechanical energy lost = 29400 J

c)

let the height is h

1000 * 9.8 * h + 29400 = 0.50 * 1000 * 2^2 + 1000 * 9.8 * 20

solving for h

h = 17.2 m

the height of the hill needed is 17.2 m

12)

let the average force is F

Using work energy theorum

m *g * h = F * d

2.25 * 9.8 * 0.30 = F * 0.0100

F = 661.5 N

average force acting on the hammer is 661.5 N

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