Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
A sledge moving over a smooth horizontal surface of ice at a velocity v2 drives out on a horizontal road and comes to a halt. The sledge has a length l, mass m and friction between runners and road is mhu
Dear studentPlease tell what is to be done in questionRegards
Assume that the spring compression is x .Here , the Initial kinetic energy of the block K.E = 1/2 mv²When the spring is compressed at it maximum range at that point the Kinetic energy will be changed to potential energy .P.E = 1/2 mx²Now from the law of conservation of energy .we get1/2 mv² = 1/2 kx²mv² = kx²v² = kx²/mx² = (v² × m) /kThe maximum compression x, is ⇒Now for the second case , we assume that there is no loss in the spring. So the kinetic energy is the same when spring comes to it's original position.But velocity is a vector quantity , so it will not be same as like original . Hope it Helps.
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -