Request a call back

Join NOW to get access to exclusive study material for best results

JEE Class main Answered

Please solve question no.7
question image
Asked by vidyavikram10 | 05 Aug, 2019, 16:43: PM
answered-by-expert Expert Answer
Obviously, 10kg block is sliding down and 5 kg block is pulled up.
 
Forces acting on the blocks are shown in figure. T is tension force pulling the block.
friction force f1 = 0.3×10g cos37 is acting against the motion of 10 kg block as shown in figure.
friction force f2 = 0.1×5g cos53 is acting against the motion of 5 kg block as shown in figure.
( g is acceleration due to gravity )
 
Let a be the acceleration of both the blocks.
 
If we apply newtons law to 10 kg block, we get,   10g sin37 -T -0.3×10g cos37 = 10×a  ....................(1)
If we apply newtons law to 5 kg block, we get,    T -5g sin53  -0.1×5g cos53 = 5×a  ....................(2)
 
By adding eqn.(1) and (2), tension T is eliminated. Then we get,
 
10g sin37 -0.3×10g cos37 -5g sin53  -0.1×5g cos53 = 15×a ......................(3)
 
from eqn.(3), We get acceleration a = 0.95 m/s2
 
 
Answered by Thiyagarajan K | 05 Aug, 2019, 18:07: PM
JEE main - Physics
Asked by rambabunaidu4455 | 03 Oct, 2024, 16:03: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by ratchanavalli07 | 17 Sep, 2024, 07:46: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by adithireddy999 | 03 Sep, 2024, 09:35: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by vaishalinirmal739 | 29 Aug, 2024, 18:07: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by vradhysyam | 26 Aug, 2024, 17:17: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT