(14%) Problem 6: A block of mass m,-16 kg slides along a horizontal surface (wit

Question

(14%) Problem 6: A block of mass m,-16 kg slides along a horizontal surface (with friction, ,-0.39) a distance d = 2.15 m before striking a second block of mass m 6.75 kg. The first block has an initial velocity of v = 8.75 m/s Randomized Variables m1 = 16 kg m2 = 6.75 kg Pk= 0.39 d = 2.15 m v = 8.75 m/s Im d ©theexpert ta.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 8% 92% cosO cotan0 asin0 acos0 atan)acotan()sinhO Submissions Attempts remaining: 8 (4% per attempt) detailed view sin tan() 4% 4% coshO tanh) cotanhO END Degrees O Radians DEL | CLEAR Submit Hint I give up! Hints: 3% deduction per hint. Hints remaining: 5 Feedback: 4% deduction per feedback 50% Part (b) How far does block two travel, d2 in meters, before coming to rest after the collision?

Explanation / Answer

v0 = 8.75 m/s

and a = - uk g = - 0.39 x 9.81 = - 3.83 m/s^2

Applying vf^2 - v0^2 = 2 a d

v^2 - 8.75^2 = 2(-3.83)(2.15)

v = 7.75 m/s ....this is speed of m1 just before collision.

Applying momentum conservation for the collsiion,

(16 x 7.75) + 0 = 0 + (6.75 v2)

v2 = 18.4 m/s ........Ans(A)

(b) Applying 0^2 - 18.4^2 = 2(-3.83) d

d = 44.1 m

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