mass of block ma= 20 kg 1) Find the speed of the 26 kg block just before it hits

Question

mass of block ma= 20 kg

1) Find the speed of the 26 kg block just before it hits the ledge?

2)Find the angular speed of the pulley at that time?

3)Find the tensions in the strings (left one in newtons)?

4)Find the tensions in the strings (right one in newtons)?

5)Find the time it takes for the 26 kg block to reach the ledge?

Flipit Physics-20174 Fall,F - Home work 2 130 Exam flipitphysics.com Course/ViewProblem?unititen10-242702&enrollmentiD.265562; Tipler6 9.P.076. The system shown below is released from rest. The m 26 kg block is 2 m above the ledge. The pulley is a uniform disk with a radius of 10 cm and mass m 4kg Assume that the string does not stip on the pulley. 1) (a) Find the speed of the 26 kg block just before it hits the ledge 1.22625 You currently have 1 submissions for this question. Only 2 submission are allowed. You can make t more submissions for this question. m/s Submit Your submissions: 1.22625 X Computed value: 1,22625 Feedback: Submitted: Friday, November 17 at 5:24 PM 4 5 6 8 2

Explanation / Answer

Writing force equation on blocks , m1 is mb, m2 is 20 kg, m is mo

m1g - T1 = m1a

T2 - m2g = m2a ,

now writing torque equation on pully,

(T1-T2)r = i alpha where i = 0.5mr^2 and alpha = a/r

T1-T2 = 0.5 ma

adding this to first two equation,

m1g - m2g = (m1+m2+0.5m)a

a = (m1-m2)*g/ (m1+m2+0.5m) = [26-20]*9.8/[26+20+0.5*4] = 1.225 m/s^2

now using third equation of motion, v = sqrt(2as) = sqrt(2*1.225*2) = 2.2136 m/s answer

2] angular speed w = v/r = 2.2136/0.10 = 22.136 rad/s answer

3] Tleft = T2 = m2g + m2a = 20*(9.8+1.225) = 220.5 N answer

4] Tright = T1 = m2g -m2a = 26 * (9.8 - 1.225) = 222.95 N answer

5] t = sqrt(2h/a) = sqrt(2*2/1.225) = 1.807 s answer

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