bhb help PH 201 Section 102 Test 1 Page 3 Problem 2 (20 points) F=12N A 0.95 kg

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PH 201 Section 102 Test 1 Page 3 Problem 2 (20 points) F=12N A 0.95 kg block is supported by a rough plane inclined at an angle of 28° with respect to the horizontal. This block is pulled up the incline by a 12.0 N force oriented at an angle of 36 with respect to the inclined plane (see the Figure). Let the (efficient of static friction be 0.74 and the coefficient of kinetic friction be 0.42. Calculste the acceleration of the block 289 Torcess-01 3) 4) kinetic fictw (0.1510.2) Fret body dagrovm

Explanation / Answer

Fx = 12*cos36 = 9.71 N

Fy = 12*sin36 = 7.05 N

along y axis

Fy + FN = m*g*cos28

normal force = FN = m*g*cos28 - Fy = (0.95*9.8*cos28)-(7.05) = 1.17 N

frictional force fk = uk*FN = 0.42 *1.17 = 0.49 N

Fnet = Fx-fk = 9.71-0.49 = 9.22 N

but Fnet = m*a

a = Fnet/m = 9.71 m/s^2 <----answer

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