Quiz 5: Chap 8 and 7. Begin Date: 4/9/2018 11:00:00 AM -- Due Date: 4/13/2018 9:

Question

Quiz 5: Chap 8 and 7. Begin Date: 4/9/2018 11:00:00 AM -- Due Date: 4/13/2018 9:00:00 AM End Date: 4/13/2018 9:00:00 AM (50%) Problem 2: A block of mass m ,-19.5 kg slides along a horizontal surface (with friction, uk 0.24) a distance d2.05 m before striking a second block of mass m2 -5.5 kg. The first block has an initial velocity of v 6.75 m/s Randomized Variables mi19.5 kg m2 = 5.5 kg 0.24 d2.05 m v-6.75 m/s ©theexpertta.com 50% Part (a) Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s? Grade Summary Deductions Potential V2 0% 100% sinO cotan atan) acotanO s cosh)tanhO cos0 Submissions Attempts remaining: 7 tan() ? ( acos) sinh) sin 4 5 6 90 per attempt) detailed view 1 2 3 cotnhO 0 Degrees O Radians Submit Hint Hints: 0% deduction per hint. Hints remaining: 5 Feedback: 0% deduction per feedback. 50% Part (b) How far does block two travel, d2 in meters, before coming to rest after the collision?

Explanation / Answer

Work done by the frictional force = change in kinetic energy of m1

-mu_k*m1*g*S = 0.5*m1*(V^2-u^2)

-0.24*19.5*9.8*2.05 = 0.5*19.5*(v^2-6.75^2)

v = 6 m/s

after impact

using

law of conservation of linear momentum

pi = pf

(m1*v) = m2*v2

(19.5*6) = (5.5*v2)

v2 = 21.3 m/s

b) Wf = -0.5*m2*v2^2

-mu_k*m2*g*S = -0.5*m2*v2^2

-0.24*5.5*9.8*S = -0.5*5.5*21.3^2

S = 96.44 m

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