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

If we use the Newton's law for Tension then

we can write mg-T=2a

where T=3.4a

therefore mg-3.4a=2a

or 2*9.8=5.4a

or a=19.6/5.4=3.629m/s^2

now when m1 leves the edge of table d is covered

using V^2=U^2+2aS we get

V=sqrt(2*3.629*.980)=2.66m/s [taking U=0

b) the kinetic energy of m1=1/2m1V^2=1/2*3.4*2.66^2= 12.02Joules

c) The speed is considered V1 then S=1.28m-.980=.3m

the using V1^2=U^2+2aS we get

V1^2=2.66^2+2*3.629*.3

or V1=3.04m/s

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