Two blocks are connected by a string, The smooth inclined surface makes an angle of 35 degrees with the horizontal, and the block on the incline has a mass of 5.7kg. The mass of the hanging block is m=3.2kg. Find the direction and magnitude of the hanging blocks acceleration?
Given
theta = 35 deg
m of incline = 5.7 kg
m of hanging block = 3.2 jg
a = ?
Can we use this equation
f = mgsin(theta)
Henry is wrong... lol
Well, you certainly can use that equation, but I have another equation for you to consider. It's called "Funnier's Law of Humorous Physics." According to this revolutionary theory, the direction and magnitude of the hanging block's acceleration can be determined by the amount of laughter it generates. The more laughter it elicits, the stronger its acceleration!
So instead of focusing on boring equations, let's focus on coming up with some hilarious jokes to make that hanging block accelerate like a rocket! Ready? Here we go:
Why don't scientists trust atoms?
Because they make up everything!
Why did the scarecrow win an award?
Because he was outstanding in his field!
Okay, okay, I know it's not a conventional way of solving physics problems, but laughter is always the best medicine, right? And who knows, maybe all that laughing will create enough positive energy to push that hanging block in the right direction.
Yes, we can use the equation f = m * g * sin(theta) to find the force component acting down the incline. In this case, the force is due to the gravitational pull on the block.
To find the direction and magnitude of the hanging block's acceleration, we need to consider the forces acting on the system. We have the force of gravity acting vertically downwards and the force along the incline due to the tension in the string.
Let's begin by finding the force component acting down the incline. The formula to calculate this force is:
f = m * g * sin(theta)
Where:
f = force acting down the incline
m = mass of the block on the incline (5.7 kg)
g = acceleration due to gravity (9.8 m/s^2)
theta = angle of the incline (35 degrees)
Plugging in the values, we have:
f = 5.7 kg * 9.8 m/s^2 * sin(35 degrees)
Now, we need to find the net force acting on the system. The net force is equal to the mass of the hanging block (m = 3.2 kg) multiplied by its acceleration (a). Using Newton's second law (F = m * a), we can express the net force as:
f = m * a
Since the hanging block's acceleration is in the same direction as the force acting down the incline, we can say that the net force is equal to the force down the incline. Therefore, we can set f = 5.7 kg * 9.8 m/s^2 * sin(35 degrees), and solve for a.
a = (5.7 kg * 9.8 m/s^2 * sin(35 degrees)) / 3.2 kg
Evaluating this expression will give us the magnitude and direction (positive or negative) of the hanging block's acceleration.
W1 = mg = 5.7kg * 9.8N/kg = 55.9 N. =
Wt of block 1.
F1 = 55.9N @ 35 Deg.
Fp = 55.9*sin35 = 32 N. = Force parallel to incline.
a = Fp/m2 = 32 / 3.2 = 10 m/s^2, Up.