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In the given figure, the natural length of spring is 0.4 m and spring constant is 200 N/m. The 3kg slider and attached spring are released from rest at end move in the vertical plane. The slider comes in rest at the point B. The work done by the friction during motion of slider is |
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Answer» Answer: bead of mass m can slide without friction on a fixed circular horizontal RING of radius 3R having centre at the point C. The bead is attached to one of the ends of spring of spring constant k. Natural LENGTH of spring is R and the other end of the spring is fixed at point O as shown in figure. Bead is released from position A, what will be kinetic energy of the bead when it reaches at point B? question
December 27, 2019avatar Preethiga Sajeev SHARE ANSWER Given, mass of bead =m radius of circular horizontal ring =3R natural length of spring =R According to conservation law of energy, KE i
+PE i
=KE f
+PE f
⇒0+ 2 1
k[OA−R] 2 =KE f
+ 2 1
k[OB−R] 2
⇒ 2 1
−k[5R−R] 2 =KE f
+ 2 1
k[R−R] 2
⇒KE f
=8kR 2 . solution December 27, 2019 avatar Toppr SHARE Practice IMPORTANT Questions UPTU 2017 Physics 50 Qs Solve related Questions System shown in the figure is released from rest. Pulley and spring is massless and friction is absent EVERYWHERE. The speed of 5 kg block when 2kg block leaves the CONTACT with ground is (take force constant of string = K=40 N/m and g=10 m/s 2 ): 1 Verified Answer In the diagram shown, the blocks A and B are of the same mass M and the mass of the block C is M 1
. Friction is present only under the block A. The whole system is suddenly released from the state of rest. The minimum coefficient of friction to keep the block A in the state of rest is equal to 1 Verified Answer |
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