A body of mass 0.5 kg is placed on the ground the coefficient of friction between the mass is and the ground is 0.2 the horizontal force f equals to 2 sin 30 applied on the Mars as shown in figure find the velocity of the body when acceleration become zero 4 first time after start
A body of mass 0.5 kg is placed on the ground the coefficient of friction between the mass is and the ground is 0.2 the horizontal force f equals to 2 sin 30 applied on the Mars as shown in figure find the velocity of the body when acceleration become zero 4 first time after start (B) g = 0.4 0.8 (D) is not defined A monkey B shown in figure is holding on to the tail of monkey A who is climbing up a rope. The masses of the monkey A and B are 5 kg and 2 kg, respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on rope to carry the monkey B with it? (A) 70 N V*-getween 70 and 105 N (B) 105 N (D) cannot say A body of mass 0.5 kg is placed on ground. The coefficient of friction between the mass and the ground is = = 0.2. A horizontal force F = (2 sin t) is applied on the mass as shown in the figure. Find the velocity of the body when its acceleration becomes zero for the first time aner start. (Take g = 10 rn/s2) (C) 2.73 m/s (B) 3.6 m/s (D) none of these Two blocks A and B are separated by some distance and mel kg tied by a string as shown in the figure. The force of friction F' = 2 t 2 Sint 0.5 kg ny2kg F.5N 0.5 in both the blocks at t = 2 sec is Two blocks A and B of same mass are connected through a Aring and arranged as shown in figure. When the system is released from rest and there is no friction, then 0.6
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Answered by
1
Answer:
Dear Student,
net force = F-f = 2sin t – f
a=0
=>(2sint – f )/0.5=0
=>sint = ½
=>t= π/6
a=4sint-0.2
v=∫adt
=-4cost-0.2t| (from 0 to π/6)
=-2√3-0.2π/6+4
=0.43 m/s
Answered by
2
Answer:
Dear Student,
net force = F-f = 2sin t – f
a=0
=>(2sint – f )/0.5=0
=>sint = ½
=>t= π/6
a=4sint-0.2
v=∫adt
=-4cost-0.2t| (from 0 to π/6)
=-2√3-0.2π/6+4
=0.43 m/s
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