Question

Conservation of Energy Roller Coaster A roller coaster cart of mass 100kg travels on a track with one loop. Fill in blanks A-H. А. KE=OJ PE=120000J h= А. V= B B KE=___CE PE=60000J h= _D V= E KE=__F PE=40000J h=__G_ V= KE= PE= h=Om v= K D E F G H K​

Answer 1

(a) The height of the roller coaster at 120,000 potential energy is 122.45 m.

(b) The velocity of the roller coaster at 0 J kinetic energy is 0.

(c) The height of the roller coaster at 60,000 potential energy is 61.23 m.

(d) The velocity of the roller coaster at 60,000 J kinetic energy is 34.64 m/s.

(e) The height of the roller coaster at 40,000 potential energy is 40.82 m.

(f) The velocity of the roller coaster at 80,000 J kinetic energy is 40 m/s.

The given parameters:

• mass of the roller coaster, m = 100 kg

When the kinetic energy = 0 and potential energy = 120,000 J

The height of the roller coaster is calculated as follows;

P.E = mgh

$h = \frac{P.E}{mg}\\\\h = \frac{120,000}{100 \times 9.8} \\\\h = 122.45 \ m$

Since the kinetic energy = 0, the velocity of the roller coaster = 0

When the potential energy, P.E = 60,000 J, the kinetic energy, K.E is calculated as;

P.E + K.E = M.A

P.E + K.E = 120,000

60,000 + K.E = 120,000

K.E = 120,000 - 60,000

K.E = 60,000 J

The height of the roller coaster at 60,000 potential energy is calculated as follows;

$h = \frac{P.E}{mg} \\\\h = \frac{60,000}{100 \times 9.8} \\\\h =61.23 \ m$

The velocity of the roller coaster at 60,000 J kinetic energy is calculated as follows;

$K.E = \frac{1}{2} mv^2\\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{ \frac{2K.E}{m}} \\\\v = \sqrt{ \frac{2\times 60,000}{100}}\\\\v = 34.64 \ m/s$

When the potential energy, P.E = 40,000 J, the kinetic energy, K.E is calculated as;

P.E + K.E = M.A

40,000 + K.E = 120,000

K.E = 120,000 - 40,000

K.E = 80,000

The height of the roller coaster at 40,000 potential energy is calculated as follows;

$h = \frac{P.E}{mg} \\\\h = \frac{40,000}{100 \times 9.8} \\\\h = 40.82 \ m$

The velocity of the roller coaster at 80,000 J kinetic energy is calculated as follows;

$v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 80,000}{100} } \\\\v = 40 \ m/s$

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