20:51 physics247com iPad ? 18kg 1.2kg 2. A 10.0 kg mass, m, on a frictionless ta

Question

20:51 physics247com iPad ? 18kg 1.2kg 2. A 10.0 kg mass, m, on a frictionless table is accelerated by a 5o kg mass, m, hanging over the edge of the table. What is the acceleration of the mass along the table? 3. Two identical blocks are tied together with a string which passes over a pulley at the crest of the inclined planes, one of which makes an angle o, - 28° to the horizontal, the other makes the complementary angle 62. If there is no friction anywhere, with what acceleration do the blocks move? A 2.00 kg and a 6.00 kg block are connected by a light string over a frictionless pulley. The two blocks are allowed to move on a fixed s wedge inclined 30°, - o.18 Determine the acceleration of the two blocks and the tension in the string. 5. Two blocks are moving down a ramp that is inclined at 30°. Box 1(of mass 1.55 kg) is above box 2(of 3.1 kg) and they are connected by a rod having negligible mass. Find the tension of the rod if the coefficient of kinetic friction for box 1 is 226 and box 2 is.113. 0.80 k 6. Two blocks, one o.8 kg and the other 2.o kg are connected by a massless string over a frictionless pulley. The coefficient of Okg kinetic friction is o.14, and the downward ramp angle is 60 degrees. 60° a) Determine the acceleration of the blocks, b) Calculate the tension of the string. A s.okg mass is accelerated from rest at the bottom of the 4.0 m long ramp by a falling 20.okg mass suspended over a frictionless

Explanation / Answer

3. block that rest on incline of 28 deg.

Applying Fnet = m a

T - m g sin28 = m a ..... (i)

for block that rest on 62 deg incline,

mg sin62 - T = m a . ....(ii)

(i) + (ii) => m g sin62 - m g sin28 = 2 m a

a = (g / 2) (sin62 - sin28)

a = 2.03 m/s^2 ............Ans

9. On m1:

m1 g sin20 - T = m1 a ...... (i)

On m2:

T - m2 g sin20 = m2 a ...... (ii)

(i) +(ii) => m1 g sin20 - m2 g sin20 = (m1+ m2) a

a = (9.81 sin20) (20 - 10) / (20 + 10)

a = 1.12 m/s^2 ........Ans

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