Question

Question 8 III Not yet answered For an object moving in uniform circular motion, the direction of the instantaneous acceleration vector is. Marked out of 2.00 Prag question Select one: O a directed radialy outward Ob tangent to the path of motion O c directed radially inward O d. equal to zero Question 9 Not yet answered Marted out of 3.00 Pag Question The crate has a mass of 80 kg and is being towed by a chain which is always directed at 20° from the horizontal as shown. If the magnitude of P is increased until the crate begins to slide, determine the crate's initial acceleration if the coefficient of static friction is 4-05 and the coefficient of kinetic friction is 03. Select one O a 353 m/s2 Ob.33.3 m/s2 None Od 036 m/s O 1.66 ms2

Answer 1

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1 for uniform circular motion- raking the tangential component of velocity is constant Hence, the tangential acceleration is zero, due to no velocity change 2.0) But the radial component of acceleration exists because of centripetal force ie muro cabang 2.9 4 I centri entripetal force Ans: directed radially inward. 'Centre of rotation. olay FBD:- N(Normal reaction) psin 20 P 120° Peos 20 (friction) F = MN mg(weight) for static equilibrium: Ms=0.5. Balancing forces a and a direction. Pcos 20= MN P MN y direction y direction N+ Psin 20 = mg N= mg- Psinzo cos 20 j cos20 :. P= u(mg-Psin 20) P=0.5 (80*9.81 – Psin 20) co20 P= 0.532 (784.8) - P(0.182) ; P=0.532 (784.8) 1.182 P-353.23N

When the box is in motion Me changes to MK = 0.3 :. Using Newton's second of law in horizontal direction or reclin? Pcos20 MN 80 a : 353.23 cos 20 0.31 mg - Psin 20) 80 a = 353.23 cos 20 - 0.3(80*9.81-353:23 Sin 20) Q-- 1.659 mis 2 Ans: 1.66 m/s2 ma - !