14.
The 40 kg block is moving to the right with a speed of 1.5 m/s when it is acted upon by the forces F1
and F2. These forces vary in the manner shown in the graph. Find the velocity (in m/s) of the block at
t = 12 s. Neglect friction and masses of the pulleys and cords.
Attachments:
Answers
Answered by
8
Answer:
velocity will be 12 m/s.
Explanation:
To get the velocity of the block, from the graph we can get that the impulse acting on the block is equal to the area under the graph.
So, we will get that m1v1 + integration of Fxdt will be equal to the m2v2 from there on substituting the values from the question we will get the value of v2 as follow:-
40(1.5) + 4[(30)4 + 10(6 - 4)] - [10(2) + 20(4 - 2) + 40(6 - 4)] = 40v2
Now, v2 on solving we will get the value as 12 m/s
Similar questions