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

1. Feb 10, 2008

### deanr

1. The problem statement, all variables and given/known data

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

2. Relevant equations

T1+U1->2=T2

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

3. 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 im working out U1->2 incorrectly

2. Feb 10, 2008

### 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..