A 2.0 kg block is placed against a spring on a frictionless 30 degrees incline (from the horizontal). The spring, whose spring constant is 1960 N/m, is compressed 20 cm and then released. How far along the incline does it send the block? (Ans.= 4m) PHYS2425.S02A.ERichards Xckboardcdn.com/5d4c11057ee7b/2177806?X-Blackboard-Expiration=1614308400000&X-Blackboard-Signature=WqB..3 /5 | -118%Problem: A 2.0 kg block is placed against a spring on a frictionless 30 incline (from thehorizontal). The spring, whose spring constant is 1960 N/m, is compressed 20 cm and thenreleased. How far along the incline does it send the block? (ANS. = 4 m)(A) PictorialRepresentation(B) PhysicalRepresentation(C) MathRepresentationand SolutionM

A 2.0 kg block is placed against a spring on a frictionless 30 degrees incline (from the horizontal). The spring, whose spring constant is 1960 N/m, is compressed 20 cm and then released. How far along the incline does it send the block?

A 2.0 kg block is placed against a spring on a frictionless 30 degrees incline (from the horizontal). The spring, whose spring constant is 1960 N/m, is compressed 20 cm and then released. How far along the incline does it send the block?

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Question

PHYS2425.S02A.ERichards X
ckboardcdn.com/5d4c11057ee7b/2177806?X-Blackboard-Expiration=1614308400000&X-Blackboard-Signature=WqB..
3 /5 | -
118%
Problem: A 2.0 kg block is placed against a spring on a frictionless 30 incline (from the
horizontal). The spring, whose spring constant is 1960 N/m, is compressed 20 cm and then
released. How far along the incline does it send the block? (ANS. = 4 m)
(A) Pictorial
Representation
(B) Physical
Representation
(C) Math
Representation
and Solution
M

Expert Solution

Step 1

$G i v e n t h a t, A 2.0 k g b l o c k, m = 2.0 k g 30 d e g r e e s i n c l i n e (f r o m t h e h o r i z o n t a l) s p r i n g c o n s \tan t i s k = 1960 N / m c o m p r e s s e d 20 c m a n d t h e n r e l e a s e d, x = 20 c m W e k n o w t h a t, t h e e l a s t i c p o t e n t i a l e n e r g y o f t h e c o m p r e s s e d s p r i n g i s, U = \frac{1}{2} k x^{2} = \frac{1}{2} \times (1960 N / m) \times {(20 c m \times (\frac{1.0 m}{100 c m}))}^{2} = 39.2 J$

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