Solving for the Final Speed of a Falling Sphere in a Pulley System

[fluffy91]

Homework Statement

A 4kg Ball and a 3kg block of wood were placed in an arrangement as shown below. When the ball was released it fell through the distance of 5m pulling the block of wood up the ramp. If the pulleys were well oiled and the surface of the ramp exerts a frictional force of 2N. What is the final speed of the sphere after falling through the 5m?

Homework Equations

E=MGH
work done= Distance moved x force
1/2mv² =K.E

The Attempt at a Solution

P.E= MGH = 4x10x5 =200 J
energy used to raised the block = 3x3x10+ (2x4) =98J
K.E of sphere after 5 m fall = 200-98 =102J
1/2mv² =102J
V=7.14m/s

But then answer is 5.34m/s

How to solve tis question. I have attached the diagram below

http://img400.imageshack.us/img400/3519/77494753vj1.jpg

 

[Oerg]

it seems you have forgotten about the frictional force. You should take the work done by the frictional force into consideration.

 

[Oerg]

You will also have to consider the addition of both masses in the kinematic equations at the end

 

[fluffy91]

You will also have to consider the addition of both masses in the kinematic equations at the end

What do u mean by addition of both masses. I have taken the frictional force into account in the (2x4)

 

[fluffy91]

Is tis correct i still have to minus the resultant force times the total distance traveled by both of them like root 3² +4² =5 then 5x9=45

1/2mv² =102-45
v = 5.34m/s

but i do not understand why i have to minus away the 45

 

[Oerg]

The 3kg load moves through a distance of 5m, not 4m. The work done by the frictional force shoulb be 2x5.

there are 2 methods of doing this, if we use the conversation of energy method like you tried, shouldn't the equation be:

P.E. lost by 4kg mass=W.D. by frictional force+P.E. gained by 3kg mass+K.E. gained by both masses

Now you need only to find out the K.E. gained by both masses.

Notice that the 2 objects move through the same distance and have the same final velocity. From there, you should be able to find the final velocity.

 

[fluffy91]

The 3kg load moves through a distance of 5m, not 4m. The work done by the frictional force shoulb be 2x5.

there are 2 methods of doing this, if we use the conversation of energy method like you tried, shouldn't the equation be:

P.E. lost by 4kg mass=W.D. by frictional force+P.E. gained by 3kg mass+K.E. gained by both masses

Now you need only to find out the K.E. gained by both masses.

Notice that the 2 objects move through the same distance and have the same final velocity. From there, you should be able to find the final velocity.

What i dun understand is why do they have the same final velocity?

 

[Oerg]

Let's imagine 2 boxes laid to rest on a frictionless surface. They are tied together by a string. If pull the first box along with a force of 2 N, wouldn't both box accelerate at the same speed and have the same final velocity? Its the same for this question. The 4kg ball is pulling the box along.

 

[fluffy91]

Let's imagine 2 boxes laid to rest on a frictionless surface. They are tied together by a string. If pull the first box along with a force of 2 N, wouldn't both box accelerate at the same speed and have the same final velocity? Its the same for this question. The 4kg ball is pulling the box along.

But F=MA so if the mass is heavier won't it accelerate less than the lighter mass? so won't the final velocity be different?

 

[fluffy91]

So won't the velocity be different?? since both mass are diff

 

[Oerg]

nope, the velocities and acceleration of both objects will be the same. In fact, this is the condition to solve such "pulling a box tied to a box" kind of question. The driving force per object will thus be proportionate to the mass of the object since acceleration is the same for both objects given that there is no friction.