Physics homework?? I'm lost!!?
A 2200 g mass starts from rest and slides a distance ℓ down a frictionless 27◦ incline, where it contacts an unstressed 120 cm long spring of negligible mass as shown in the figure. The mass slides an additional 17 cm as it is brought momentarily to rest by compressing the spring of force constant 21 N/cm. The acceleration of gravity is 9.8 m/s^2. Note: The spring lies along the surface of the ramp (see figure). Assume: The ramp is frictionless. Now, the external force is rapidly removed so that the compressed spring can push up the mass.
Find the initial separation ℓ between mass and spring. Answer in units of m.
I've already tried 1.41, and it was wrong. Can someone help me solve this with a good explanation please? Thank you!
- NCSLv 72 years agoFavorite Answer
Initial GPE becomes SPE:
mgh = ½kx²
2.200kg * 9.8m/s² * (ℓ + 0.17m) * sin27º = ½ * 2100N/m * (0.17m)²
ℓ = 2.93 m
which ought to be rounded to 2.9 m owing to the data.
Hope this helps!
- electron1Lv 72 years ago
I suggest that all the distances be converted from centimeters to meters.
Spring constant = 2100 N/m
The following force causes the object to accelerate.
Force parallel = 2.2 * 9.8 * sin 27 = 21.56 * sin 27
Let’s use following equation to determine the work that is done on the spring.
Work = ½ * k * d^2
d is the distance the spring is compressed.
d = 0.17 meter
Work = ½ * 2100 * 0.17^2 = 30.345 N * m
This is equal to the object’s kinetic energy just before it hit the spring.
KE = ½ * 2.2 * v^2 = 1.1 * v^2
1.1 * v^2 = 30.345
v^2 = 30.345 ÷ 1.1
v = √(30.345 ÷ 1.1)
The object’s velocity just before it hit the spring is approximately 5.25 m/s. This is its maximum velocity. Let’s use the following equation to determine the distance it slides.
vf^2 = vi^2 + 2 * a * d, vi = 0, a = 9.8 * sin 27
30.345 ÷ 1.1= 2 * 9.8 * sin 27 * d
d = (30.345 ÷ 1.1) ÷ (19.6 * sin 27)
This is approximately 3.1 meters. To determine the distance l, add 0.17meter.I hope this is correct.