Tired of walking up the stairs, an 80kg engineering student designs an ingenious

Question

Tired of walking up the stairs, an 80kg engineering student designs an ingenious device for reaching his third floor dorm room. A block is attached to a rope that passes over a pulley. The student holds the other end of the rope. When the block is released, the student is pulled up to his dorm room, 8m off the ground, in a time of 1.8s.

I need to ultimately find the mass of the block, and the acceleration.

So far i have:

mass of student = m_s=80kg

mass of block = m_b

g = 9.8m/s^2

a = acceleration

Force due to tension in rope= F_T

So, then:

Sum of forces on student = F_T-m_s*g=m_s*a

Sum of forces on block = F_T-m_b*g=m_b*a

Solving for F_T, I got:

m_s*a+m_s*g=m_b*a+m_b*g

But then I am not really sure how to find the mass of the block, or what steps to really take next. Help???

Explanation / Answer

h = at^2/2

8 = a x 1.8^2/2

a = 2.47 m/s2

on student : T - 80g = 80 x 2.47

T = 982.4 N

on mass:

mg - T = m x 2.47

m (g - 2.47) = 982.4

m =133.84 kg

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