Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping
Answer:
Explanation:
Let assume begins movement at zero point, that is, height is equal to zero. The block has an initial linear kinetic energy and no gravitational potential energy and end with no linear kinetic energy, some gravitational potential energy and work losses due to slide friction. In mathematical terms, this system can be model as follows:
Where 
are linear kinetic energy, gravitational potential energy and work, respectively.
<h2>
Answer: 12 s</h2>
Explanation:
The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.
In this sense, the main movement equation in the Y axis is:
(1)
Where:
is the instrument's final position
is the instrument's initial position
is the instrument's initial velocity
is the time the parabolic movement lasts
is the acceleration due to gravity at the surface of planet X.
As we know and 
when the object hits the ground, equation (1) is rewritten as:
(2)
Finding :
(3)
(4)
(5)
Finally:
