Total distance = 36500 m
The average velocity = 19.73 m/s
<h3>Further explanation</h3>Given
vo=initial velocity=0(from rest)
a=acceleration= 1 m/s²
t₁ = 20 s
t₂ = 0.5 hr = 1800 s
t₃= 30 s
Required
Total distance
Solution
State 1 : acceleration
State 2 : constant speed
State 3 : deceleration
Total distance : state 1+ state 2+state 3
the average velocity = total distance : total time
Answer:
0.075 m
Explanation:
The picture of the problem is missing: find it in attachment.
At first, block A is released at a distance of
h = 0.75 m
above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:
where
is the acceleration due to gravity
is the mass of the block
is the speed of the block A just before touching block B
Solving for the speed,
Then, block A collides with block B. The coefficient of restitution in the collision is given by:
where:
e = 0.7 is the coefficient of restitution in this case
is the final velocity of block B
is the final velocity of block A
is the initial velocity of block B
Solving,
Re-arranging it,
(1)
Also, the total momentum must be conserved, so we can write:
where
And substituting (1) and all the other values,
This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to
where
k = 600 N/m is the spring constant
x is the compression of the spring
And solving for x,
