R KH Ox=1,2m \"m:20 kg kommunerna k=2soon MMF 01 5) JLI is IL, maximum hi by u b

Question

R KH Ox=1,2m "m:20 kg kommunerna k=2soon MMF 01 5) JLI is IL, maximum hi by u box? 13 k=1500x d= 0.8m. m=20 imi* * This bir3 Mn-O.G The box has initially a height of 3m. After the box is released from rest it will slide down the incline onto a level track which has a rough surface (u=0.4) of length 0.8m. and a spring with k=250ON/m. Find the maximum height reached by the box after it returns to the incline. 4) For problem 3, what is the velocity at the bottom of the incline after the initial release, and what is the velocity at the bottom of the incline after the block bounced at the spring and returned?

Explanation / Answer

3) velocity at the bottom of the slide after it's released from 3m

mgh = 1/2 mv^2

(9.8) ( 3) =1/2 v^2

v^2=58.8

we will again use conservation of energy to find the compression of spring

1/2 mv^2 = 1/2kx^2 + mu ( mgx)

0.5 ( 20 ) 58.8 = 0.5 (2500) x^2 + 0.4 ( 20) (9.8) (0.8)

588- 62.72= 1250x^2

x =0.648 m

Let the max height reached after returning to incline = h

1/2kx^2 = mgh +  0.4 ( 20) (9.8) (0.8)

0.5 (2500) (0.4199) =(20) (9,8) h + 62.72

h = 2.358 m apprx

Q.4: velocity at bottom after initial release =sqroot ( 58.8) =7.668 m/s

velocity at bottom of slide after rebouncing from spring

1/2 v^ 2= g( 2.358)

v = 6.798 m/s

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