Answer:
v = 0.028 m/s
Explanation:
In this case you take into account that all the elastic potential of the box, when the spring is compressed is equal to the kinetic energy of the box before it hits the spring. That is:
(1)
m: mass of the box = 4.5kg
k: spring constant = 85N/m
x: compression of the spring = 6.5cm = 0.0065m
You solve the equation (1) for v:
(2)
Next, you replace the values of the parameters:
The velocity of the box when it hits the spring is 0.028m/s
Answer:
Kinetic energy of the projectile at the vertex of the trajectory: .
Work done when firing this projectile: .
Explanation:
Since the drag on this projectile is negligible, the horizontal velocity of this projectile would stay the same (at
) throughout the flight.
The vertical velocity of this projectile would be
at the vertex (highest point) of its trajectory. (Otherwise, if
, this projectile would continue moving up and reach an even higher point. If
, the projectile would be moving downwards, meaning that its previous location was higher than the current one.)
Overall, the velocity of this projectile would be when it is at the top of the trajectory. The kinetic energy
of this projectile (mass
) at the vertex of its trajectory would be:
.
Apply the Pythagorean Theorem to find the initial speed of this projectile:
.
Hence, the initial kinetic energy of this projectile would be:
.
All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be .