A 1.30-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed of v, = 2.60 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Figure e). (a) Find the distance compression d (in m). 0.5 x You will find that you need to use the quadratic equation to find this distance. m (b) Find the speed v (in m/s) at the unstretched position when the object is moving to the left (Figure d). 2.44 How much energy is stored in the spring just before the object makes contact with the spring? How much energy Jx stored in the spring after the object leaves the spring? Do you need to consider the spring potential energy working this part? m/s (c) Find the distance D (in m) where the object comes to rest. 0.5 This part is easiest to work by starting with your answer to part (b) (if it is correct!), but it can also be worked using only your answer to part (a). To understand how, think about the initial state of the system (object traveling with a speed 2.60, spring uncompressed) and the final state of the system (object at rest, spring uncompressed). What happens to the kinetic energy of the object? m

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A 1.30-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed of v, = 2.60 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes
to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a
distance D to the left of the unstretched spring (Figure e).
v= 0
0 =A
e
(a) Find the distance of compression d (in m).
0.5
You will find that you need to use the quadratic equation to find this distance. m
(b) Find the speed v (in m/s) at the unstretched position when the object is moving to the left (Figure d).
2.44
How much energy is stored in the spring just before the object makes contact with the spring? How much energy is stored in the spring after the object leaves the spring? Do you need to consider the spring potential energy in working this part? m/s
(c) Find the distance D (in m) where the object comes to rest.
0.5
This part is easiest to work by starting with your answer to part (b) (if it is correct!), but it can also be worked using only your answer to part (a). To understand how, think about the initial state of the system (object traveling with a speed 2.60,
spring uncompressed) and the final state of the system (object at rest, spring uncompressed). What happens to the kinetic energy of the object? m
(d) What If? If the object becomes attached securely to the end of the spring when it makes contact, what is the new value of the distance D (in m) at which the object will come to rest after moving to the left?
1.22
m

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Step 1

Since you have posted a question with a multiple sub parts, we will solve first three sub - parts for you. To get the remaining sub parts solved please repost the complete question and mention the sub parts to be solved.

$m a s s = 1.3 k g μ = 0.25 v_{i} = 2.6 m . s k = 50 N / m a) u \sin g c o n s e r v a t i o n o f e n e r g y i n i t i a l k i n e t i c e n e r g y = e n e r g y l o s t t o f r i c t i o n + s p r i n g p o t e n t i a l e n e r g y (1 / 2) (1.3) * {(2.6)}^{2} = (0.25) (1.3) (9.8) (d) + (1 / 2) (50) {(d)}^{2} 25 d^{2} + 3.185 d - 4.394 = 0 o n s o l v i n g t h e q u a d r a t i c e q u a t i o n d = 0.36 m$

Step 2

$b) u \sin g c o n s e r v a t i o n o f e n e r g y (1 / 2) (50) {(0.36)}^{2} = (1 / 2) (1.3) {(v)}^{2} + (0.25) (1.3) (9.8) (0.36) 3.24 = 0.65 v^{2} + 1.1466 v = 1.8 m / s$

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