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
Michael
Dec 5, 2015

(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.

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