A string that does not stretch connects two blocks, m_1=3.40kg (originally at re

Question

A string that does not stretch connects two blocks, m_1=3.40kg (originally at rest) on a horizontal table at a height 1.28m above the floor, to m_2=2kg originally a distance d=0.980m above the floor and also initially at rest. The surface of the table and its edge are frictionless. The sliding block m_1 continues to move horizontally after reaching the edge of the table. The hanging block m_2 stops immediately without bouncing when it hits the floor. a.) Using Conservation of Energy, find the speed at which m_1 leaves the edge of the table. (Assume m_2 hits the ground before m_1 leaves the table.) b.) What is the Kinetic Energy of m_1 just before it leaves the table? c.) Find the speed of m_1 just before it hits the floor.

Explanation / Answer

a)

All the PE stored in the system must be converted to KE of the system

So, m2*g*d = 0.5*(m1+m2)*v^2

So, 2*9.8*0.98 = 0.5*(2 + 3.4)*v^2

So, v = 2.67 m/s <----------answer

b)

KE of m1 = 0.5*m1*v^2

= 0.5*3.4*2.67^2 = 12.1 J

c)

All the PE of the mass m1 is converted to KE

So, initial PE + initial KE = final KE

So, mgh + 0.5*mv^2 = 0.5*m*v'^2

So, v' = sqrt(2gh + v^2) = sqrt(2*9.8*1.28 + 2.67^2)

= 5.68 m/s <--------answer

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