Two blocks with masses m1 = 3.00 kg and m2 = 5.00 kg are connected by a light string that slides over two frictionless pulleys as shown.Initially m2 is held 5.00 m off the floor while m1 is on the floor. The system is then released. ?

a) At what speed does m2 hit the floor?
b) Find the speed of m1 when it is at height of 1.80 m? c) What is the maximum height attained by m1?

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1 Answer
Dec 5, 2015

Answer:

(a)

#4.95"m/s"#

(b)

#2.97"m/s"#

(c)

#5"m"#

Explanation:

#(a)#

Mass #m_2# experiences #5g"N"# downwards and #3g"N"# upwards giving a net force of #2g"N"# downwards.

The masses are connected so we can regard them as acting as a single 8kg mass.

Since #F=ma# we can write:

#2g=(5+3)a#

#:.a=(2g)/8=2.45"m/s"^(2)#

If you like to learn formulae the expression for 2 connected masses in a pulley system like this is:

#a=((m_2-m_1)g)/((m_1+m_2))#

Now we can use the equations of motion since we know the acceleration of the system #a#.

So we can get the speed that #m_2# hits the ground #rArr#

#v^2=u^2+2as#

#v^2=0+2xx2.45xx5#

#v^2=24.5#

#:.v=4.95"m/s"#

#(b)#

#v^2=u^2+2as#

#:.v^2=0+2xx2.45xx1.8#

#v^2=8.82#

#:.v=2.97"m/s"#

(c)

Since #m_2# cannot drop any further than 5m I reason that #m_1# cannot go any higher than 5m.