Help mechanics

For the first one, the string goes slack in the subsequent motion so a=-9.8 (assuming upwards is postive). Also the initial velocity is 1.2. So get "s" and add 0.8 which is the initial displacement from the floor when the string goes slack. Obviously, there are a couple of mistakes in what you did.

c) and d) are similar for the 2nd one. A diagram helps, but when the 3kg mass hits the floor, there is no tension as the string goes slack and the only force (deceleration) on the 2kg mass is due to the rough surface. So similar to the previous question. For d) draw the diagram with tensions on and hence there must be a reaction from the pulley to balance the resultant force due to the tensions. Using pythagoras you can almost write the answer down based on part a).

(edited 11 months ago)

Reply 2

OP

Original post by mqb2766

For the first one, the string goes slack in the subsequent motion so a=-9.8 (assuming upwards is postive). Also the initial velocity is 1.2. So get "s" and add 0.8 which is the initial displacement from the floor when the string goes slack. Obviously, there are a couple of mistakes in what you did.

c) and d) are similar for the 2nd one. A diagram helps, but when the 3kg mass hits the floor, there is no tension as the string goes slack and the only force (deceleration) on the 2kg mass is due to the rough surface. So similar to the previous question. For d) draw the diagram with tensions on and hence there must be a reaction from the pulley to balance the resultant force due to the tensions. Using pythagoras you can almost write the answer down based on part a).

I don’t get it still… look what I’ve gathered for the second Q part C… I got a to be 0.4m :frown:

Reply 3

Similar mistakes, wrong acceleration and forgetting to add on the initial displacement. A sketch really should help. The particle has travelled 1m when the string goes slack, so your 0.4 is half the requried 0.8. The deceleration is due to the rough table/friction, not gravity.

Reply 4

OP

Original post by mqb2766

Similar mistakes, wrong acceleration and forgetting to add on the initial displacement. A sketch really should help. The particle has travelled 1m when the string goes slack, so your 0.4 is half the requried 0.8. The deceleration is due to the rough table/friction, not gravity.

What’s the acceleration?

Reply 5

The last sentence tells you what the force due to resistance is (10N), and hence the decleration, acting on the 2kg mass is. In the previous question you seem to have underlined the important parts and translated them to a sketch. Thats good practice to hopefully make sure you dont miss anything.

Reply 6

OP

Original post by mqb2766

The last sentence tells you what the force due to resistance is (10N), and hence the decleration, acting on the 2kg mass is. In the previous question you seem to have underlined the important parts and translated them to a sketch. Thats good practice to hopefully make sure you dont miss anything.

Hey I’m struggling so much and my exams are in 2 weeks finals!
look I’ve attempted the question but can’t do it. I learn by looking at displayed solutions can u plz send me a worked solution to this

Reply 7

f = ma
-10 = 2a
a = -5
Its a 10 N force acting in the opposite direction to the motion so its negative.

Youve not marked the 10N force on your diagram and the diagram is wrong for this question part as there is no tension (string slack) and the 3kg mass has hit the floor. Nor is the initial displacement marked on there.

(edited 11 months ago)

Reply 8

OP

Original post by mqb2766

f = ma
-10 = 2a
a = -5
Its a 10 N force acting in the opposite direction to the motion so its negative.

Youve not marked the 10N force on your diagram and the diagram is wrong for this question part as there is no tension (string slack) and the 3kg mass has hit the floor. Nor is the initial displacement marked on there.

So it’s like this

can u plz show me how to draw it plzzzz I’m so stressed, just send a pic of the diagram you draw here. Also why do u add 1 to 0.8… because it’s for the 2kg particle and the 1m is above ground for the 3kg

Reply 9

Agreed for that part, its 0.8m displacement from when the string goes slack/the 3kg mass hits the ground. Both masses have moved 1m from their original initial position so the 2kg mass has moved 1.8m in total.

Its a fairly trivial "diagram" for c), its a 2kg mass moving horizontally with a resistive/re tarding force of 10N agianst the direction of motion. The initial displacement is 1m and the initial velocity is 2.8m/s. Thats it, however if you dont get this right/clear, its likely youll be applying suvat incorrectly. Hence why you do sketches to translate the info in the question into a form youre clear about.

In your diagram, the 3kg mass/pulley/tension and even the 2g weight are all irrelevant for this part.

Reply 10

OP

Original post by mqb2766

Agreed for that part, its 0.8m displacement from when the string goes slack/the 3kg mass hits the ground. Both masses have moved 1m from their original initial position so the 2kg mass has moved 1.8m in total.

Its a fairly trivial "diagram" for c), its a 2kg mass moving horizontally with a resistive/re tarding force of 10N agianst the direction of motion. The initial displacement is 1m and the initial velocity is 2.8m/s. Thats it, however if you dont get this right/clear, its likely youll be applying suvat incorrectly. Hence why you do sketches to translate the info in the question into a form youre clear about.

In your diagram, the 3kg mass/pulley/tension and even the 2g weight are all irrelevant for this part.

Ah I see thanks!!! Also can u explain part d as I’m crying rn

Reply 11