A string going over a massless frictionless pulley connects two blocks of masses 5.9 kg and 9.5 kg. the 5.9 kg block lies on a 31 ◦ incline; the coefficient of kinetic friction between the block and the incline is µ = 0.22. The 9.5 kg block is hanging in the air. The 9.5 kg block accelerates downward while the 5.9 kg block goes up the incline with the same acceleration.Given g = 9.8 m/s^2 what is the acceleration
of the system?
Fn = m1gcos31
Ff = mu Fn
Fnet = m2g - Ff - m1gsin31
Fnet = ma where m is total mass of system
To find the acceleration of the system, we need to analyze the forces acting on each block separately.
First, let's consider the 5.9 kg block on the incline. We need to find the force of friction acting on it. The force of friction can be calculated using the equation:
F_friction = µ * N
where µ is the coefficient of kinetic friction and N is the normal force. The normal force is the force exerted by the incline on the block perpendicular to the incline's surface. In this case, it is equal to the component of the gravitational force acting perpendicular to the incline.
N = mg * cos(θ)
where m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of the incline (31 degrees).
Next, we find the gravitational force component acting parallel to the incline. This force is equal to:
mg * sin(θ)
Since the block is moving up the incline, the force of friction opposes the motion and acts in the opposite direction. Therefore, the net force acting on the block is given by:
F_net = mg * sin(θ) - F_friction
Now, we can apply Newton's second law to the 5.9 kg block:
F_net = m * a
where m is the mass of the block and a is the acceleration.
Simplifying the equation, we have:
mg * sin(θ) - µ * mg * cos(θ) = m * a
Now, let's move on to the 9.5 kg hanging block. The only significant force acting on it is the gravitational force pulling it downward, given by:
F_gravity = m * g
Since the two blocks are connected by a string passing over a pulley, they have the same acceleration and the force acting on the hanging block is equal to the tension in the string. Therefore:
T = m * a
We can now set up a system of equations by equating the expressions for tension in the string:
T = m * a
And balancing the forces on the 5.9 kg block:
mg * sin(θ) - µ * mg * cos(θ) = m * a
Since we want to find the acceleration of the system, we can eliminate T from the equations:
m * a = m * g * sin(θ) - µ * m * g * cos(θ)
Simplifying further by dividing through by m:
a = g * (sin(θ) - µ * cos(θ))
Plugging in the provided values:
a = 9.8 * (sin(31) - 0.22 * cos(31))
Evaluating this expression, we find that the acceleration of the system is approximately 5.51 m/s^2.