- Thread starter innovas1
- Start date

- Status
- Not open for further replies.

what is the velocity?? I say its 4, my bro says it is 2 and argued with me. Should we lump the 3 masses together to calculate velocity or just calculate vel using the 2kg mass?

some say vel is 2, some say 4...

wat is yr view? ANy physics teacher or expert here?

innovas1 said:

what is the velocity?? I say its 4, my bro says it is 2 and argued with me. Should we lump the 3 masses together to calculate velocity or just calculate vel using the 2kg mass?

some say vel is 2, some say 4...

wat is yr view? ANy physics teacher or expert here?

innovas1 said:

if we just consider the mass on the table, then the tension to the left is 18N and 10N to the right. Resultant force of 8N acts on the 2kg mass. so, a=4,v=4 at that instant.

here is the answer if you look at the breakdown of forces. red is forces acting on the individual objects (minus the gravity and reactional forces), blue is the resultant force on each object. thus it's clear to see that the answer is 2m/s for velocity. as to how do you get the breakdown of the forces shown, think about it slowly, heh.

wacko said:

here is the answer if you look at the breakdown of forces. red is forces acting on the individual objects (minus the gravity and reactional forces), blue is the resultant force on each object. thus it's clear to see that the answer is 2m/s for velocity. as to how do you get the breakdown of the forces shown, think about it slowly, heh.

wacko said:

here is the answer if you look at the breakdown of forces. red is forces acting on the individual objects (minus the gravity and reactional forces), blue is the resultant force on each object. thus it's clear to see that the answer is 2m/s for velocity. as to how do you get the breakdown of the forces shown, think about it slowly, heh.

Assume the whole set up is in equibrum...

Thus, before 8N force is applied, tension in both ropes are 9.81N.

When 8N force is applied, looking at the FBD of the 2kg mass. We'll have, 8+9.81N acting on the left and 9.81N still acting on the right... Therefore, the resultant force on the 2kg mass is still 8N to the left. (The 9.81N cancel out each other)

Thus overall resultant force acting on the 2kg mass is 8N to the right... We all know that F=ma. Therefore, at 1s the acceration is thus 8/2=4m/s^2. Velocity also 4m/s

After 1s. When 8N is taken away, applying newton's law of motion, assuming zero frictional force. We'll have Acceration=0, velocity=4m/s.

My 2 cent's worth...

Kira said:

Okie my version...

Assume the whole set up is in equibrum...

Thus, before 8N force is applied, tension in both ropes are 9.81N.

When 8N force is applied, looking at the FBD of the 2kg mass. We'll have, 8+9.81N acting on the left and 9.81N still acting on the right... Therefore, the resultant force on the 2kg mass is still 8N to the left. (The 9.81N cancel out each other)

Thus overall resultant force acting on the 2kg mass is 8N to the right... We all know that F=ma. Therefore, at 1s the acceration is thus 8/2=4m/s^2. Velocity also 4m/s

After 1s. When 8N is taken away, applying newton's law of motion, assuming zero frictional force. We'll have Acceration=0, velocity=4m/s.

My 2 cent's worth...

Assume the whole set up is in equibrum...

Thus, before 8N force is applied, tension in both ropes are 9.81N.

When 8N force is applied, looking at the FBD of the 2kg mass. We'll have, 8+9.81N acting on the left and 9.81N still acting on the right... Therefore, the resultant force on the 2kg mass is still 8N to the left. (The 9.81N cancel out each other)

Thus overall resultant force acting on the 2kg mass is 8N to the right... We all know that F=ma. Therefore, at 1s the acceration is thus 8/2=4m/s^2. Velocity also 4m/s

After 1s. When 8N is taken away, applying newton's law of motion, assuming zero frictional force. We'll have Acceration=0, velocity=4m/s.

My 2 cent's worth...

wacko said:

erroneous assuption that the tension in the string is 17.81N. let's put it this way, if the tension IS 17.81N, then the FBD of the block P will have resultant force of 0N, and if there is zero resultant force, why is the block P accelerating in the first second?

Kira said:

17.81N acts to the left. 9.81N acts to the right.... Resultant force is 8N acting to the left. Thus block move left...

bearycute said:

er... u r right that the resultant force is 8... but that is the resultant force on the system which consist of 3 masses adding to 4kg... thus accln is 2...hope that is clear... guess i might have to draw free body diagram on each of the masses to show if still not clear...

:dunno:

Kira said:

But doesn't the 1kg mass cancel out each other? i.e. P aids motion, Q resist motion?????

:dunno:

:dunno:

W the application of a 8N force on the system ( which consist of 3 masses), a net force results. Applying newton's second law: F = ma, the resultant force F = 8N. m = (1+2+1)= 4 kg. Thus the accln is 2. all the masses share the same accln = 2 for the 1st second in which the force is applied.

v = u + at. Since initial speed is zero, final speed v = 0 + 2t => v = 2 since t = 1. After one second when the 8N force is removed, the system moves at constant speed as again now the net force acting on the system is zero(newton's 1st law) ignoring all resistive forces....

so when the force of 8N is being applied.... it can be thought of juz as a 8N force pulling on the trolley........ (the downward forces that are contributed by P and Q can be ignored because both are acting downwards = in the opp. position so in the end they cancel out each other)

using formula v=u+at

u=0

t=2

F=MA

8=19.62(A)

A= 8m/s^2

therefore, v= 16m/s

since this is a MCQ.....

can post the options out ?

- Status
- Not open for further replies.