What is the acceleration of the block?

Question

A block of mass 3 kg slides down an inclined plane at an angle of 45° with a massless tether attached to a pulley with mass 1 kg and radius 0.5 m at the top of the incline (see the following figure). The pulley can be approximated as a disk. The coefficient of kinetic friction on the plane is 0.4. What is the acceleration of the block?

Answer 1

Athena, I don't see the picture. Let's assume it is understandable.

We know that angle Θ = 45° , mass of the block m₁ = 3 kg, mass of the pulley m₂ = 1 kg,

radius of the pulley R = 0.5 m, coefficient of friction µ = 0.4

There are 4 forces acting on block of mass m₁:

m₁g - weight, N - normal reaction force, T - tension, and f - frictional force.

Newton's second law for the block in scalar form is:

m₁gsinΘ - T - f = m₁a

Because f = μN and N = m₁gcosΘ , after substitution we will get:

m₁sinΘ - T - μm₁gcosΘ = m₁a. (1)

Newton's second law for rotation of the pulley has a form: Iα = TR.

Moment of inertia for the disk I = 1/2 m₂R² , angular acceleration of the pulley α = a/R , where a is linear acceleration and R is the radius of the pulley. Hence, we can write:

TR = 1/2 m₂R² a/R or T = 1/2 m₂a (2)

Solving together equations (1) and (2), we will find out that acceleration

a = [m₁g(sinΘ - μcosΘ)]/(m₁ + 1/2m₂)

Hope that now you can plug in the numbers by yourself and get the numerical value for the acceleration.

Good luck to you!