A small block of mass m1 = 0.500 kg slides to the left, with a velocity of v1 =

Question

A small block of mass m1 = 0.500 kg slides to the left, with a velocity of v1 = ?8.40 m/s, towards a curved wedge of mass m2 = 3.00 kg which sits on an icy, frictionless lake surface. When the block reaches the wedge, it slides up to a height 3.00 m above the ice. It comes to a stop relative to the wedge, and the two objects slide together to the left. (a) What is the velocity v2 of the wedge-block system when they are sliding together? (b) Kinetic energy is not conserved during this process. What is the change in kinetic energy?

Explanation / Answer

the block is the only provider of energy to the block wedge for the system initially the block is moving with speed v1 = 8.40 mts/sec the blocks initial K.E is used by block finally to reach a height h above the wedge and set the wedge into motion => 1/2m1(v1)^2 = m1gh + 1/2(m1 + m2)(v2)^2 substituting v2 = 1.5 mts/sec final K.E of block wedge sys is 1/2(3.5)(1.5)^2 = 3.93J initial K.E of block is 1/2*0.5*(8.4)^2 = 17.64 J change in K.E is 13.71J

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