Block 1, of mass m1 = 0.650kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. For an angle of θ = 30.0∘ and a coefficient of kinetic friction between block 2 and the plane of μ = 0.400, an acceleration of magnitude a = 0.600m/s2 is observed for block 2.
Part A
Find the mass of block 2, m2.
they why do you [post if you don't have the answer!^^^
Kinda funny how some of these questions get answered years later
this was kool!
Well, if the mass of block 2 is a mystery, then we'll have to get to the bottom of it! But don't worry, I won't make any weight jokes.
Let's start by writing down the equations of motion for the system. We'll consider the forces acting on block 2. We have the gravitational force pulling the block downwards, the tension in the string pulling it upwards, and the frictional force opposing its motion.
Since the block is moving horizontally, we'll focus on the forces in that direction. The gravitational force component and the tension in the string will cancel each other out, leaving us with the frictional force as the net force.
The equation for the frictional force is given by f_friction = μ * normal_force, where μ is the coefficient of kinetic friction and the normal force is equal to the gravitational force acting on the block.
Now, we know that the net force is equal to mass times acceleration. So we can write:
f_friction = m2 * a
Substituting the expression for the frictional force, we get:
μ * m2 * g = m2 * a
Canceling out m2 on both sides and solving for m2, we find:
m2 = μ * g / a
Just plug in the given values for μ, g, and a, and you'll get your answer. Don't forget to convert the angle to radians if necessary!
Now, if only my calculations had such mass-terful precision!
To find the mass of block 2, m2, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
In this case, the force acting on block 2 is the force of gravity pulling it downwards, which is equal to the product of its mass, m2, and the acceleration due to gravity, g (9.8 m/s^2):
Force of gravity on block 2 = m2 * g
However, there is an additional force acting on block 2 due to the friction between block 2 and the plane. The frictional force can be calculated as the product of the coefficient of kinetic friction, μ, and the normal force, which is equal to the force of gravity acting on block 2:
Frictional force on block 2 = μ * (Force of gravity on block 2)
Since the block is accelerating, the net force on block 2 can be calculated as:
Net force on block 2 = (mass of block 2) * (acceleration of block 2)
Now, we can equate the net force and the sum of the forces acting on the block:
Net force on block 2 = Force of gravity on block 2 - Frictional force on block 2
(mass of block 2) * (acceleration of block 2) = m2 * g - μ * (Force of gravity on block 2)
Substituting the given values, we have:
(mass of block 2) * (0.600 m/s^2) = m2 * (9.8 m/s^2) - 0.400 * (m2 * 9.8 m/s^2)
Simplifying the equation, we get:
0.600 m2 = 9.8 m2 - 3.92 m2
Rearranging the equation, we get:
0.600 m2 + 3.92 m2 = 9.8 m2
4.52 m2 = 9.8 m2
Dividing both sides of the equation by 4.52, we find:
m2 ≈ 2.17 kg
Therefore, the mass of block 2, m2, is approximately 2.17 kg.
(m1)(g – a) = (m2)(a + gsin(θ) + μgcos(θ))
(0.650)(9.80-0.600)=(m2)(0.600+9.8*sin(30)+0.400*9.80*cos(30))
m2=0.672301423 kg
three sig figs---->m2=0.672 kg