In order to be safe going around a loop the centripetal acceleration cannot exceed (see problems from yesterday).

in order to be safe the centripetal acceleration should be higher than 39.2 m/s²

Explanation:

Assuming no friction while travel around the loop then we can apply the conservation of energy

where

K = kinetic energy = 1/2*m*v²

and m= mass , v= velocity at the bottom

K = kinetic energy = m*g*h , where g= gravity and h= height from the bottom of the loop

then by conservation of energy

Energy in the bottom of the loop = K + V(=0 because of h=0) = K = (1/2)*m*v²

Energy in the top of the loop = K(=0 in the limiting case of v=0 , that is falling from the loop) + V= V = m*g* 2R

where R= radius of the loop

then

Energy in the bottom of the loop=  Energy in the top of the loop

(1/2)*m*v²= m*g* 2R

(1/2)*v²= 2*g* R

v²/R = 4*g

the centripetal acceleration ac is defined as

ac= v²/R= 4*g

then since g=9.8 m/s²

ac= 4*9.8 m/s² = 39.2 m/s²

therefore in order to be safe the centripetal acceleration should be higher than 39.2 m/s²

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.

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.