W = F*dAnswer: Work Done by Force F on Cylinder Mass m1: 352.2 J

[aarce]

A cylinder of mass m1 = 30 kg and radius r = 8 cm lies on a board of mass m2 = 60 kg. The ground is frictionless and the coefficient of friction (both static and kinetic) between the board and the cylinder is u = 0.1. The centre of mass of the cylinder is pulled with a force of F = 44.15 for two seconds. Find the work done by force F.
Step 1: I drew a picture

Step 2: determined my knowns which is mass 1= 30 kg, radius = 8 cm, mass 2 = 60, u = 0.1, F= 44.15 for t = 2 sec.

Step 3: T.V.= F

Step 4: Equation: ?

 

[I like Serena]

A cylinder of mass m1 = 30 kg and radius r = 8 cm lies on a board of mass m2 = 60 kg. The ground is frictionless and the coefficient of friction (both static and kinetic) between the board and the cylinder is u = 0.1. The centre of mass of the cylinder is pulled with a force of F = 44.15 for two seconds. Find the work done by force F.
Step 1: I drew a picture

Step 2: determined my knowns which is mass 1= 30 kg, radius = 8 cm, mass 2 = 60, u = 0.1, F= 44.15 for t = 2 sec.

Step 3: T.V.= F

Step 4: Equation: ?

Hi aarce! Welcome to MHB! ;)

Let's start with the applicable equations.
The force of friction has a maximum of:
$$F_{f,max} = \mu F_n$$
where $F_n$ is the normal force given by:
$$F_n = m_1 \cdot g$$

The net forces on $m_1$ and $m_2$ are:
$$F_1 = F - F_f$$
$$F_2 = F_f$$

If we assume that the masses start sliding with respect to each other, we have that $F_f = F_{f,max}$.
What will be the respective accelerations then?