The initial speed of the bolt is not 58.86 m/s.
Let a be the acceleration of the rocket.
During the 4 sec lift off, the rocket has reached a height of
h = (1/2)*a*t^2
with t=4,
h = (1/2)*a^16
h = 8*a
Its velocity at 4 sec is
v = t*a
v = 4*a
The initial velocity of the bolt is thus 4*a.
During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,
h = (1/2)*g*t^2 + V0*t
Substituting h0=8*a, t=6 and V0=-4*a into it,
8*a = (1/2)*g*36 - 4*a*6
Solving for a
a = 5.52 m/s^2
Answer:
a2 = 2.5 m/s2
Explanation:
F1 = m1 a1 We use the same force so F1 = F2
= 5kg × 15m/s2 F2 = m2 a2
= 75N a2 is required
a2 = F2 / m2
= 75N / 30 kg
= 2.5 m/s2
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
