Answer:
v = 0.489 m/s
Explanation:
It is given that,
Mass of a box, m = 1.5 kg
The compression in the spring, x = 6.5 cm = 0.065 m
Let the spring constant of the spring is 85 N/m
We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.
So, the speed of the box is 0.489 m/s.
Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:
where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,
<u>h = 0.82 m</u>
Now, for the time in air during upward motion we use first equation of motion:
(c)
Now we will consider the downward motion and use the third equation of motion:
where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,
<u>vf = 7.17 m/s</u>
Now, for the time in air during downward motion we use the first equation of motion:
(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
<u>t = 1.14 s</u>
The equal and opposite force is the force of the stone's gravity pulling up on thy Earth.
<u>Explanation</u>: