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Two blocks of masses 5 kg and 2 kg (see in the fig. I`ve attached) are initially at rest on floor. They are connected by a light string, passing over a frictionless pulley and maintained constant. Calculate acceleration a1 and a2 of the 5 kg and 2 kg masses, respectively when F is 110 N {g=10m/s^2}

Two blocks of masses 5 kg and 2 kg (see in the fig. I`ve attached) are initially at rest on floor. They are connected by a light string, passing over a frictionless pulley and maintained constant. Calculate acceleration a1 and a2 of the 5 kg and 2 kg masses, respectively when F is 110 N {g=10m/s^2}

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Grade:11

2 Answers

Arun
25750 Points
6 years ago
Dear Piyush
 
When F = 110 N
110 - 2T = 0
T = 55 N
Thus both the weights will be lifted up
55 -50 = 5 * a1
a1 = 1 m/sec²
55 -20 = 2 * a2
a2 = 17.5 m/sec²
 
Regards
Arun (askIITians forum expert)
Harsh
16 Points
6 years ago
Assuming all The strings and pulleys are massless ...Next you need to understand that for the masses to be even lifted there is a certain force required ... The lighter block will move first when force is applied and heavier block will not move ..For the lighter block to move Force > = tension T (tension) > m1g .... (1)T > 2g ..... (2) From 1 and 2 Force = m1 × 2g = 2 ×2g = 40N Thus masses will move only when force is greater than 40N As force is 110N we can apply Newton law...Tension is equally distributed so force = 2 × tension Tension in each string 55 N T - m2g = m2a255- (5×10)=5a1 a1 =1ms^2T-m1g= m1a155-(2×10)=m2a2a2 = 17.5ms^2 So a1= 1ms ^2 and a2 =17.5 ms^2

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