Aroller coaster car of mass 1420 kg slides on a frictionless track starting at a distance 36.83 m above the bottom of a loop 29 m in diameter. the acceleration of gravity is 9.81 m/s 2. if friction is negligible, what is the magnitude of the force of the track on the car when the car is at the top of the loop? answer in units of n.

Magnitude of force = 6415.207 N

Explanation:

Given that

m= 1420 kg

h= 36.83 m

d=29 m ,r=14.5 m

g=9.81 m/s²          

From energy conservation

Lets take velocity of particle at top position is V

m g h  - m g d =1/2 m V²    

V² = g(h-d)

V²  = 9.81 x (36.83 - 29)

V=8.76 m/s

At top position

F  + m g = mV² /r

By putting the values

F=−6415.207 N

Magnitude of force = 6415.207 N

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.

what is the actual question though?