Find the velocity of a block + 2 cylinders acted on by a force for 2meteres

[deanr]

Homework Statement

A 50 kg block is transported a short distance by using two cylindrical rollers, each having a mass of 17.5 kg. If a horizontal force P=125N is applied to the block, determine the block's speed after it has been displaced 2m to the left. Originally the block is at rest. No slipping occurs.

The radius of the cylinders: .5m

Homework Equations

T1+U1->2=T2

U=F*d+M*$$\theta$$ ?

The Attempt at a Solution

$$\omega = 2 \times v$$

Icyl=2.1875 kgm^2
T1=0
$$U1->2=125 \times2+(125 \times .5) \times 2$$
$$T2(block)=\frac {1}{2} \times m \times v^{2}$$
$$T2(rollers)=2 \times ( \frac{1}{2} \times m \times v ^{2})+2 \times ( \frac{1}{2} \times Icyl \times (2v)^{2})$$

the answer should be v=2.8 m/s but I think I am working out U1->2 incorrectly

 

[Littlepig]

Well, you know that the force that is acting on the rollers is equal to the force applied contrary to the P, as there's no slliping...

so, the aceleration of the block is a=(P-2T)/m where m is block mass, P the force, and T the force that provokes a torque on each roller..

 

[Moetasim]

. Can someone please help me with this problem?

I would first start by identifying the key variables and equations needed to solve this problem. The key variables are the masses of the block and cylinders, the applied force, the displacement, and the radius of the cylinders. The key equations are the work-energy theorem (T1+U1->2=T2) and the equations for rotational kinetic energy (T2(block) and T2(rollers)).

Next, I would draw a diagram of the situation to help visualize the problem and to identify any assumptions that need to be made. In this case, we can assume that there is no slipping between the block and the cylinders, and that the force is applied horizontally.

Then, I would set up the equations and plug in the given values to solve for the final velocity of the block after it has been displaced 2m to the left. The work-energy theorem can be written as T1+U1->2=T2 where T1 is the initial kinetic energy, U1->2 is the work done by the force, and T2 is the final kinetic energy. In this case, T1=0 since the block is initially at rest. U1->2 is equal to the work done by the force, which is given by F*d, where F is the applied force and d is the displacement. T2 can be calculated using the equation T2=\frac{1}{2}mv^2, where m is the mass of the block and v is the final velocity.

For the rotational kinetic energy of the block, we can use the equation T2(block)=\frac{1}{2}I\omega^2, where I is the moment of inertia of the block and \omega is the angular velocity. Since the block is moving in a straight line, we can use the relationship \omega=\frac{v}{r}, where r is the radius of the cylinders. For the rotational kinetic energy of the cylinders, we can use the equation T2(rollers)=\frac{1}{2}I\omega^2, where I is the moment of inertia of the cylinders and \omega is the angular velocity. In this case, the cylinders are rotating twice as fast as the block, so we can use the relationship \omega=2\frac{v}{r}.

Plugging in the given values and solving for v, we get v=2.8 m/s