Asked by Serena
A block of mass 11.2 kg slides from rest down a frictionless 24.0° incline and is stopped by a strong spring with a spring constant 31.3 kN/m (note the unit). The block slides 8.90 m from the point of release to the point where it comes to rest against the spring. When the block comes to rest momentarily, how far has the spring been compressed? Answer in units of cm and use 9.8 m/s2 for g.
I can't really figure out how to start to try to solve this problem. Help please?
I can't really figure out how to start to try to solve this problem. Help please?
Answers
Answered by
Henry
Fb = m*g = 11.2kg * 9.8N/kg = 109.8 N. =
Force of block.
Fp = 109.8*sin24 = 44.64 N. = Force parallel to the incline.
d = (1m/31,300N) * Fp
d = (1m/31,300N) * 44.64N = 0.0014262 m.
= 0.143 cm.
Force of block.
Fp = 109.8*sin24 = 44.64 N. = Force parallel to the incline.
d = (1m/31,300N) * Fp
d = (1m/31,300N) * 44.64N = 0.0014262 m.
= 0.143 cm.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.