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
A. I just took a quiz on it and A was my answer
The final velocity of the projectile when it strikes the ground below is 198.51 m/s.
<h3>
Time of motion of the projectile</h3>
The time taken for the projectile to fall to the ground is calculated as follows;
h = vt + ¹/₂gt²
where;
- h is height of the cliff
- v is velocity
- t is time of motion
265 = (185 x sin45)t + (0.5)(9.8)t²
265 = 130.8t + 4.9t²
4.9t² + 130.8t - 265 = 0
solve the quadratic equation using formula method,
t = 1.89 s
<h3>Final velocity of the projectile</h3>
vyf = vyi + gt
where;
- vyf is the final vertical velocity
- vyi is initial vertical velocity
vyf = (185 x sin45) + (9.8 x 1.89)
vyf = 149.322 m/s
vxf = vxi
where;
- vxf is the final horizontal velocity
- vxi is the initial horizontal velocity
vxf = 185 x cos(45)
vxf = 130.8 m/s
vf = √(vyf² + vxf²)
where;
- vf is the speed of the projectile when it strikes the ground below
vf = √(149.322² + 130.8²)
vf = 198.51 m/s
Learn more about final velocity here: brainly.com/question/6504879
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