# M2 Problem

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

This discussion is closed.

A block of mass 5kg is on a plane which is at an angle of 40 degrees to the horizontal. The block is connected by light,inextinsable strings to obejcts A and D which hang vertically and have masses of 5kg and 15kg respectively, as shown in the diagram on the right. Between B and C, the strings are parallel to the line of the greatest slope of the inclined plane and pass over smooth pegs B and C. The object D, which slides in a groove, is initially 1.5m above the floor. You may assume that A never reachers the peg B and that the block never reachers the peg C.

i) Calculate the change in gravitational potential energy of the whole system when moves from its initial position to the floor. Your answer should specify whether this is a loss or a gain.

The coefficient of friction between the block and the plane is 0.6. The object at A hands freely but the obect at D slides in the groove against a constant frictional force of 15N. You may assume that air resistance is negligable.

The system is released from rest from its position.

ii) Show the D hits the floor with a speed of about 2.20 ms-1

iii) After the impact of D with the floor, how much further does the block move up the plane?

i) Calculate the change in gravitational potential energy of the whole system when moves from its initial position to the floor. Your answer should specify whether this is a loss or a gain.

The coefficient of friction between the block and the plane is 0.6. The object at A hands freely but the obect at D slides in the groove against a constant frictional force of 15N. You may assume that air resistance is negligable.

The system is released from rest from its position.

ii) Show the D hits the floor with a speed of about 2.20 ms-1

iii) After the impact of D with the floor, how much further does the block move up the plane?

0

Report

#3

(Original post by

A block of mass 5kg is on a plane which is at an angle of 40 degrees to the horizontal. The block is connected by light,inextinsable strings to obejcts A and D which hang vertically and have masses of 5kg and 15kg respectively, as shown in the diagram on the right. Between B and C, the strings are parallel to the line of the greatest slope of the inclined plane and pass over smooth pegs B and C. The object D, which slides in a groove, is initially 1.5m above the floor. You may assume that A never reachers the peg B and that the block never reachers the peg C.

i) Calculate the change in gravitational potential energy of the whole system when moves from its initial position to the floor. Your answer should specify whether this is a loss or a gain.

The coefficient of friction between the block and the plane is 0.6. The object at A hands freely but the obect at D slides in the groove against a constant frictional force of 15N. You may assume that air resistance is negligable.

The system is released from rest from its position.

ii) Show the D hits the floor with a speed of about 2.20 ms-1

iii) After the impact of D with the floor, how much further does the block move up the plane?

**Ali Brown**)A block of mass 5kg is on a plane which is at an angle of 40 degrees to the horizontal. The block is connected by light,inextinsable strings to obejcts A and D which hang vertically and have masses of 5kg and 15kg respectively, as shown in the diagram on the right. Between B and C, the strings are parallel to the line of the greatest slope of the inclined plane and pass over smooth pegs B and C. The object D, which slides in a groove, is initially 1.5m above the floor. You may assume that A never reachers the peg B and that the block never reachers the peg C.

i) Calculate the change in gravitational potential energy of the whole system when moves from its initial position to the floor. Your answer should specify whether this is a loss or a gain.

The coefficient of friction between the block and the plane is 0.6. The object at A hands freely but the obect at D slides in the groove against a constant frictional force of 15N. You may assume that air resistance is negligable.

The system is released from rest from its position.

ii) Show the D hits the floor with a speed of about 2.20 ms-1

iii) After the impact of D with the floor, how much further does the block move up the plane?

Let's take D first. It's mass is 15kg, and it falls 1.5m. We take g to be 9.8 in M1 so I assume you do the same for M2. D's change in g.p.e. is given by the equation:

Δg.p.e. = mΔhg

Δg.p.e. = 15*-1.5*9.8

= -220.5J

Secondly, take A, with a mass of 5kg. The string is inextensible, and with the assumption that A never reached peg B, A travels vertically for 1.5m. using the same equation:

Δg.p.e. = 5*1.5*9.8

= 73.5J

Finally, take the block with a mass of 5kg. It travels 1.5m at an angle of 40 degrees to the horizontal upwards. Using simple trigonometry, we find it's upward displacement is 1.5 sin40. It's g.p.e. change is therefore:

Δg.p.e. = 5*1.5sin40*9.8

= 47.244...J

Keeping this number in your calculator, you can work out the total g.p.e. change by adding the three figures together.

73.5 + 47.244... - 220.5 =

**-99.76J**(2d.p.)

ii) Ok. We need to work out the velocity at which D hits the floor. That will involve Netwon's equations, and given the initial velocity and the displacement (0 and 1.5 respectively) we have only the acceleration to find. The acceleration depends on the mass of D, which is given, and the force exerted on D, which we need to find.

Since the string is inextensible, the two forces acting on D are the tension in the string, the frictional force against the groove of 15N, and D's own weight. We now need to work out the tension in the string, or the sum of the forces on the string by the block and the particle A.

Modelling these independantly:

A:

Forces:

It's weight, 5g (5kg*g), downwards

The tension in the string, directly upwards caused by D (T1)

Resolving vertically:

T1 - 5g = 5a

or

T1 = 5(a+g)

Block:

Forces:

It's weight, 5g (5kg*g), directly downwards.

The tension in the string up the plane, the force to be found, (T2)

The tension in the string down the plane caused by A, (T1)

The contact force perpendicular to the plane, (R)

Friction acting down the plane, (Fn)

Resolving perpendicular to plane:

*since there is no motion perpendicular to the plane, there must be no resultant force and we can equate the two opposing forces immediately

R = 5g cos 40

= 37.536.... N

Using the law concerning the coefficient of friction which is:

Fn = μR

We know that μ = 0.6, so we can work out the frictional force:

Fn = 0.6*5g cos40 (for accuracy, I don't write 37.54 because 5gcos40 is more accurate)

= 3cos40

Resolving paralell to the plane:

T2 - T1 - 5g sin40 - 3 cos40 = 5a

T1 and a are unknown variables, but we know T2 from modelling the forces on A. Substituting:

T2 - (5a+5g) - 5g sin40 - 3 cos40 = 5a

T2 - 5a + 5g - 5g sin40 - 3 cos40 = 5a

T2 + 5g - 5g sin40 - 3 cos40 = 0

T2 = 5g sin40 + 3 cos40 - 5g

= -15.2N

Now, modelling for D:

Forces:

It's own weight, 15g, acting downwards

Tension in the string, T2, acting upwards. This is known but I'll leave it as it's uncalculated form for accuracy.

Friction, given already as 15N, acting upwards.

F=ma

15g - |T2| - 15 = 15a

15a = 116.79...

a = 7.786... ms^-2

And in the home straight:

u = 0

v = ?

s = 1.5

a = 7.786..

v^2 = u^2 + 2as

= 0 + 2*1.5*7.786...

= 23.358...

v = 4.83 ms^-1

I am on M1, and a and b seem like very difficult versions of my syllabus. c) I'm guessing is M2 specific, and I haven't learnt it yet I don't think.

I hope this helps. I know it helped me.

0

Report

#4

(Original post by

i) Add up the gains/losses. It is immediately obvious that if the system moves with no external forces, the gravitational potential energy change is negative.

Let's take D first. It's mass is 15kg, and it falls 1.5m. We take g to be 9.8 in M1 so I assume you do the same for M2. D's change in g.p.e. is given by the equation:

Δg.p.e. = mΔhg

Δg.p.e. = 15*-1.5*9.8

= -220.5J

Secondly, take A, with a mass of 5kg. The string is inextensible, and with the assumption that A never reached peg B, A travels vertically for 1.5m. using the same equation:

Δg.p.e. = 5*1.5*9.8

= 73.5J

Finally, take the block with a mass of 5kg. It travels 1.5m at an angle of 40 degrees to the horizontal upwards. Using simple trigonometry, we find it's upward displacement is 1.5 sin40. It's g.p.e. change is therefore:

Δg.p.e. = 5*1.5sin40*9.8

= 47.244...J

Keeping this number in your calculator, you can work out the total g.p.e. change by adding the three figures together.

73.5 + 47.244... - 220.5 =

ii) Ok. We need to work out the velocity at which D hits the floor. That will involve Netwon's equations, and given the initial velocity and the displacement (0 and 1.5 respectively) we have only the acceleration to find. The acceleration depends on the mass of D, which is given, and the force exerted on D, which we need to find.

Since the string is inextensible, the two forces acting on D are the tension in the string, the frictional force against the groove of 15N, and D's own weight. We now need to work out the tension in the string, or the sum of the forces on the string by the block and the particle A.

Modelling these independantly:

A:

Forces:

It's weight, 5g (5kg*g), downwards

The tension in the string, directly upwards caused by D (T1)

Resolving vertically:

T1 - 5g = 5a

or

T1 = 5(a+g)

Block:

Forces:

It's weight, 5g (5kg*g), directly downwards.

The tension in the string up the plane, the force to be found, (T2)

The tension in the string down the plane caused by A, (T1)

The contact force perpendicular to the plane, (R)

Friction acting down the plane, (Fn)

Resolving perpendicular to plane:

*since there is no motion perpendicular to the plane, there must be no resultant force and we can equate the two opposing forces immediately

R = 5g cos 40

= 37.536.... N

Using the law concerning the coefficient of friction which is:

Fn = μR

We know that μ = 0.6, so we can work out the frictional force:

Fn = 0.6*5g cos40 (for accuracy, I don't write 37.54 because 5gcos40 is more accurate)

= 3cos40

Resolving paralell to the plane:

T2 - T1 - 5g sin40 - 3 cos40 = 5a

T1 and a are unknown variables, but we know T2 from modelling the forces on A. Substituting:

T2 - (5a+5g) - 5g sin40 - 3 cos40 = 5a

T2 - 5a + 5g - 5g sin40 - 3 cos40 = 5a

T2 + 5g - 5g sin40 - 3 cos40 = 0

T2 = 5g sin40 + 3 cos40 - 5g

= -15.2N

Now, modelling for D:

Forces:

It's own weight, 15g, acting downwards

Tension in the string, T2, acting upwards. This is known but I'll leave it as it's uncalculated form for accuracy.

Friction, given already as 15N, acting upwards.

F=ma

15g - |T2| - 15 = 15a

15a = 116.79...

a = 7.786... ms^-2

And in the home straight:

u = 0

v = ?

s = 1.5

a = 7.786..

v^2 = u^2 + 2as

= 0 + 2*1.5*7.786...

= 23.358...

v = 4.83 ms^-1

I am on M1, and a and b seem like very difficult versions of my syllabus. c) I'm guessing is M2 specific, and I haven't learnt it yet I don't think.

I hope this helps. I know it helped me.

**mik1a**)i) Add up the gains/losses. It is immediately obvious that if the system moves with no external forces, the gravitational potential energy change is negative.

Let's take D first. It's mass is 15kg, and it falls 1.5m. We take g to be 9.8 in M1 so I assume you do the same for M2. D's change in g.p.e. is given by the equation:

Δg.p.e. = mΔhg

Δg.p.e. = 15*-1.5*9.8

= -220.5J

Secondly, take A, with a mass of 5kg. The string is inextensible, and with the assumption that A never reached peg B, A travels vertically for 1.5m. using the same equation:

Δg.p.e. = 5*1.5*9.8

= 73.5J

Finally, take the block with a mass of 5kg. It travels 1.5m at an angle of 40 degrees to the horizontal upwards. Using simple trigonometry, we find it's upward displacement is 1.5 sin40. It's g.p.e. change is therefore:

Δg.p.e. = 5*1.5sin40*9.8

= 47.244...J

Keeping this number in your calculator, you can work out the total g.p.e. change by adding the three figures together.

73.5 + 47.244... - 220.5 =

**-99.76J**(2d.p.)ii) Ok. We need to work out the velocity at which D hits the floor. That will involve Netwon's equations, and given the initial velocity and the displacement (0 and 1.5 respectively) we have only the acceleration to find. The acceleration depends on the mass of D, which is given, and the force exerted on D, which we need to find.

Since the string is inextensible, the two forces acting on D are the tension in the string, the frictional force against the groove of 15N, and D's own weight. We now need to work out the tension in the string, or the sum of the forces on the string by the block and the particle A.

Modelling these independantly:

A:

Forces:

It's weight, 5g (5kg*g), downwards

The tension in the string, directly upwards caused by D (T1)

Resolving vertically:

T1 - 5g = 5a

or

T1 = 5(a+g)

Block:

Forces:

It's weight, 5g (5kg*g), directly downwards.

The tension in the string up the plane, the force to be found, (T2)

The tension in the string down the plane caused by A, (T1)

The contact force perpendicular to the plane, (R)

Friction acting down the plane, (Fn)

Resolving perpendicular to plane:

*since there is no motion perpendicular to the plane, there must be no resultant force and we can equate the two opposing forces immediately

R = 5g cos 40

= 37.536.... N

Using the law concerning the coefficient of friction which is:

Fn = μR

We know that μ = 0.6, so we can work out the frictional force:

Fn = 0.6*5g cos40 (for accuracy, I don't write 37.54 because 5gcos40 is more accurate)

= 3cos40

Resolving paralell to the plane:

T2 - T1 - 5g sin40 - 3 cos40 = 5a

T1 and a are unknown variables, but we know T2 from modelling the forces on A. Substituting:

T2 - (5a+5g) - 5g sin40 - 3 cos40 = 5a

T2 - 5a + 5g - 5g sin40 - 3 cos40 = 5a

T2 + 5g - 5g sin40 - 3 cos40 = 0

T2 = 5g sin40 + 3 cos40 - 5g

= -15.2N

Now, modelling for D:

Forces:

It's own weight, 15g, acting downwards

Tension in the string, T2, acting upwards. This is known but I'll leave it as it's uncalculated form for accuracy.

Friction, given already as 15N, acting upwards.

F=ma

15g - |T2| - 15 = 15a

15a = 116.79...

a = 7.786... ms^-2

And in the home straight:

u = 0

v = ?

s = 1.5

a = 7.786..

v^2 = u^2 + 2as

= 0 + 2*1.5*7.786...

= 23.358...

v = 4.83 ms^-1

I am on M1, and a and b seem like very difficult versions of my syllabus. c) I'm guessing is M2 specific, and I haven't learnt it yet I don't think.

I hope this helps. I know it helped me.

0

Report

#6

(Original post by

Using the law concerning the coefficient of friction which is:

Fn = μR

We know that μ = 0.6, so we can work out the frictional force:

Fn =

**mik1a**)Using the law concerning the coefficient of friction which is:

Fn = μR

We know that μ = 0.6, so we can work out the frictional force:

Fn =

**0.6*5g cos40**(for accuracy, I don't write 37.54 because 5gcos40 is more accurate)**= 3cos40**Also, I noticed that in the diagram attached, he put 4kg for A, but in the question, he wrote 5kg

0

Report

#7

If you use use mik1a's method, but with 4kg for A instead, then you get 2.16 m/s (which is approx. 2.2 m/s)

0

Report

#8

wohoo!!!

I always make silly small errors... I wish I could go through working like that and find errors like you just did, but even in exams It's such a tiring job and when I do I can't spot any errors.

Thanks, that makes me feel much better knowing my error wasn't in the method.

I always make silly small errors... I wish I could go through working like that and find errors like you just did, but even in exams It's such a tiring job and when I do I can't spot any errors.

Thanks, that makes me feel much better knowing my error wasn't in the method.

0

Report

#9

(Original post by

wohoo!!!

I always make silly small errors... I wish I could go through working like that and find errors like you just did, but even in exams It's such a tiring job and when I do I can't spot any errors.

Thanks, that makes me feel much better knowing my error wasn't in the method.

**mik1a**)wohoo!!!

I always make silly small errors... I wish I could go through working like that and find errors like you just did, but even in exams It's such a tiring job and when I do I can't spot any errors.

Thanks, that makes me feel much better knowing my error wasn't in the method.

0

Report

#11

**Ali Brown**)

A block of mass 5kg is on a plane which is at an angle of 40 degrees to the horizontal. The block is connected by light,inextinsable strings to obejcts A and D which hang vertically and have masses of 5kg and 15kg respectively, as shown in the diagram on the right. Between B and C, the strings are parallel to the line of the greatest slope of the inclined plane and pass over smooth pegs B and C. The object D, which slides in a groove, is initially 1.5m above the floor. You may assume that A never reachers the peg B and that the block never reachers the peg C.

i) Calculate the change in gravitational potential energy of the whole system when moves from its initial position to the floor. Your answer should specify whether this is a loss or a gain.

The coefficient of friction between the block and the plane is 0.6. The object at A hands freely but the obect at D slides in the groove against a constant frictional force of 15N. You may assume that air resistance is negligable.

The system is released from rest from its position.

ii) Show the D hits the floor with a speed of about 2.20 ms-1

iii) After the impact of D with the floor, how much further does the block move up the plane?

Since the string is inextensible, we know that at the point where D this the floor the speed of D is equal to the speed of the block, so we can say that u = 2.2

The question asks how far further it travels up, so the acceleration being negative will cause at some point the instantaneous velocity of the block to be 0. At this point, the block will no longer travel up the plane, so we can say v = 0

One more constant and we can find the displacement. Acceleration:

This is quite tough seeing as there is friction between the plane and the block. Resolving parallel to the plane when the string no longer provides a tension gives:

[F=ma]

5g sin 40 - Fn = 5a

Now to find Fn we must resolve perpendicular to the plane and use the equation Fn = μR:

R = 5g cos 40

Fn = μ5g cos 40

We know that μ = 0.6, and we can say that:

5a = 5g sin 40 - μ5g cos 40

So we can work out a:

a = g sin 40 - μg cos 40

= 11.22 ms^1

This is clearly wrong since and acceleration of a particle on a downward slope with firction could not possibly be greater than that on a particle in freefall!

You have to see the funny side sometimes!

0

Report

#12

(Original post by

a = g sin 40 - μg cos 40

= 11.22 ms^1

**mik1a**)a = g sin 40 - μg cos 40

= 11.22 ms^1

0

X

Page 1 of 1

Go to first unread

Skip to page:

new posts

Back

to top

to top