Question

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3 kg vertically cing is 10 kg g is used to frictionless maximum these ag is most 11. Three blocks, A, B, and C. of masses 1, 2, and 3 kg, respectively, are initially at rest on a frictionless surface as indicated in the figure above. What force has to be applied on block C 5= 200 mis? to accelerate the three blocks at 2 m/s? (A) 0.33 N (2) +22) 4362) 16 V=20mis (B) 1.5 N (C) 3.0 N rictionless (D) 6.0 N = 12 with another 12N In this case momentum 12. An electrical motor provides 0.50 W of mechanical power. How much time will it take and kinetic the motor to lift a 0.1 kg mass at constant speed from the floor to a shelf 2.0 m above the floor? ved, and (A) 0.25 s (B) 0.40 s 3, and kinetic (C) 1.0 s (D) 2.0 s ergy are 4.0 s energy is 5.0 m 2 = 0 u=0.20 at at ves along ce of a onstant. ven by height of the horizonweck and the lock slide om 13. A block slides from rest with negligible friction down the track above, descending a vertical height of 5.0 m to point P at the bottom. It then slides on the horizontal surface. The coefficient of friction between the block and the horizontal surface is 0.20. How far does the block slide on the horizontal surface before it comes to rest? (A) 0.40 m (B) 1.0 m (C) 2.5 m (D) 10 m 25 m

Answer 1

(1) Fella) +2(a) + 3(a) - bla). N given a= 2 m3 = F = 6/2) = 12N (13) using conservation of Energy, mgh I mv² : -) velocity at p} =) (9.8) (5) -IV =) V 49x2 = v= 752 m/s Let it travels a distance is' before coming to rest. here retardation, a= Mg = (0.2) 9.8 mit uring kinematic equation, tu+zas no final relouty = 0 r = 72 m/s, a=- 19,6 m52 : s= v= 49xx - 5 = 2.5m 20 2x19-6 = (12) Giren poner of moth, P = 0.5 W mass of water to be lifter, m= 0.1kg height to which water to be lifted th= am power, p = were rugh hah 0.189,8x2 - 3.92 seconds

C5 Given ret) - 3 t - 27 +4 is relocity. Acceleration, as d[vct)] ad [3ť - 2t+4] m/s² at a = 6t-2 t=2s, a = 6(2)-2 a = 10 mit . (16) relocity of particle in sm is maximum at its mean position. from the graph, upward relouty u maximum at, t=0,255, 1.255; 2,253, 3.253.. Hence from given options, 0.25s is the Answer ( Here T=15 Time period of oscillations is T= 2nl m ) K = 40m 0.4) - 15.79 Nm! = Spring Constant k = 4 (3:14) (0.4)