the two blocks of mass 2kg and 5kg are connected by an ideal string passing over a Pulley. the block of mass 2 kg is free to slide on a free slide on a surface inclined at an angle of 30°which the horizontal whereas 5 kg hangs,freely find the horizontal acerelation of the system and T in the strings. (u = 0.3)
Answers
Answered by
1
Answer:
plz provide the diagram of u have....
Answered by
0
Answer:
T=25N.
Explanation:
If we make the free body diagram of the box then we will get the component of the block from which the pulley is connected. So, if we take the masses as m1 and m2 then we will get m1g-T=m1a or 5g-5a=T where T is the tension in the string.
Again, F = μm2gcos30 which on solving we will get 0.3*2*10*√3/2 which will be 5.19 N. So, for the mass 2 kg we will get that T - 5.19 - m2gsin30=m2a which on solving we will get T-5.19-10=2a or T-15.19=2a. Now, value of the T from the 5 kg. 5g -5a -15.19=2a or 7a=50-15 which is a=35/7 which is 5 m/s^2. So, the tension will be 5*10-5*5=T or T=25N.
Similar questions