Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

auty talka g.com/ibiscms/mod/fibis/view.php?id-2195436 PHYS 10-Fal\". APRELEV Ac

ID: 1460928 • Letter: A

Question

auty talka g.com/ibiscms/mod/fibis/view.php?id-2195436 PHYS 10-Fal". APRELEV Acthities and Due Dates work and Kreti Energy 11/9/2015 11:55 PM O 41.7/10011/2/2015 0554 PM Gradeb Periodic Table Question 3 of 4 Map: sapling learning on a horizontal frictionless table. The block is pushed into the spring, compressing it by 5.00 cm, and is then released from rest. The spring begins to push the block back toward the equilibrium position at x 0 cm. The graph shows the component of the force (in N) exerted by the spring on the block versus the relative to equilibrium. Use the graph to answer the following questions. of the block (in cm) How much work is done by the spring in the block from its initial position at om to x 2.31 cm? Number F(N) Fr (N) 0 What is the speed of the block when it reaches 2.31 cm? Number m/s x (cm) What is the maximum speed of the block? Number m/s Previous eCheck AnswerNext Ex Hint

Explanation / Answer


mass of the block m=3.98 kg

comprression x=5cm

from the graph,

spring constant K=slope

K=(6-0)/(0-(-0.05))

K=120 N/m


a)

compressin x1=-5 cm

compressin x2=2.31 cm

work done W=1/2*k*(x2^2-x1^2)

work done W=1/2*120*(0.0231^2-(-0.05)^2))

w=-0.118 J


b)

let,

speed is v at x=2.32 cm

K.E=P.E

1/2*m*v^2=1/2*kx^2

1/2*3.98*v^2=1/2*120*(0.0231^2)

====> v=0.127 m/sec


c)

K.E=P.E

1/2*m*vmax^2=1/2*k*xmax^2

1/2*3.98*v^2=1/2*120*0.05^2

===> vmax=0.274 m/sec