The change in kinetic energy of the car is equivalent to the change in its potential energy. Thus:
K.E = P.E
1/2 x mΔv² = mgΔh
h = (8.2² - 5²) / 2(9.81)
h = 2.15 meters
Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
the upward force is from the table and the downward is from gravity they are equal in force so the box doesn't fly up or sink down
Answer:
(a) 1767.43 N
(b) 182.45 N
Explanation:
Radius of earth, R = 6450 km
Weight of person, W = 7070 N
mass of person, m = W / g = 7070 / 9.8 = 721.4 kg
(a) h = 6450 km
The value of acceleration due to gravity on height is given by


g' = g / 4 = 9.8 / 4 = 2.45 m/s^2
The weight of the person at such height is
W' = m x g' = 721.4 x 2.45
W' = 1767.43 N
(b) h = 33700 km
The value of acceleration due to gravity on height is given by


g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2
The weight of the person at such height is
W' = m x g'
W' = 721.4 x 0.253
W' = 182.45 N
Answer:
Kinetic energy and momentum are conserved.
Explanation: