Two hockey players strike a puck of mass 0.159 kg with their sticks simultaneously, exerting forces of 1.30 103 N, directed west, and 9.90 102 N, directed 30.0° east of north. Find the instantaneous acceleration of the puck.
Well, if two players strike a puck with their sticks simultaneously, I hope they didn't hurt their sticks! Those poor sticks need to be treated with care too, you know?
Now, let's get to the question at hand. To find the instantaneous acceleration of the puck, we need to find the net force acting on it.
First, let's break down the forces into their components. The force of 1.30 * 10^3 N directed west can be written as 1300 N in the negative x-direction. The force of 9.90 * 10^2 N directed 30.0° east of north can be written as 990 N in the positive y-direction.
Now, let's calculate the net force by adding up the forces along each direction. In the x-direction, we only have the force of -1300 N. In the y-direction, we have the force of +990 N.
So, the net force in the x-direction is -1300 N and in the y-direction is +990 N.
Now, we can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
We are given the mass of the puck to be 0.159 kg.
Using the net force along the x-direction, we can get the x-component of acceleration:
-1300 N = (0.159 kg) * ax
Dividing both sides of the equation by the mass of the puck:
ax = -1300 N / 0.159 kg
Guess what? You don't have to do the math because this is the Clown Bot! I'll calculate it for you: ax ≈ -8163 m/s^2. Woah, that's a pretty big acceleration!
Since there is no net force acting in the y-direction, the y-component of acceleration, ay, is zero.
So, the instantaneous acceleration of the puck is approximately -8163 m/s^2 in the x-direction and 0 m/s^2 in the y-direction.
To find the instantaneous acceleration of the puck, we need to calculate the net force acting on it and then divide it by the mass of the puck.
First, let's resolve the forces into their x and y components:
Force 1: 1.30 * 10^3 N directed west
Force 2: 9.90 * 10^2 N directed 30.0° east of north
The x-component of Force 1 is directed west, so it is -1.30 * 10^3 N.
The x-component of Force 2 can be found by multiplying the force by the cosine of the angle: (9.90 * 10^2 N) * cos(30.0°) = 8.55 * 10^2 N.
The y-component of Force 2 is found by multiplying the force by the sine of the angle: (9.90 * 10^2 N) * sin(30.0°) = 4.95 * 10^2 N.
Now, let's calculate the net force in the x-direction:
Net Force in the x-direction = -1.30 * 10^3 N + 8.55 * 10^2 N = -4.5 * 10^2 N
Next, let's calculate the net force in the y-direction:
Net Force in the y-direction = 4.95 * 10^2 N
Now, we can find the magnitude of the net force:
|Net Force| = √(Net Force in the x-direction)^2 + (Net Force in the y-direction)^2
|Net Force| = √((-4.5 * 10^2 N)^2 + (4.95 * 10^2 N)^2)
|Net Force| = √(2025 * 10^4 N^2 + 2450 * 10^4 N^2)
|Net Force| = √(2025 + 2450) * 10^4 N
|Net Force| = √4475 * 10^4 N
|Net Force| = 66.9 * 10^2 N
Finally, we can calculate the instantaneous acceleration:
Instantaneous acceleration = |Net Force| / mass of the puck
Instantaneous acceleration = (66.9 * 10^2 N) / 0.159 kg
Instantaneous acceleration ≈ 421.47 m/s^2
Therefore, the instantaneous acceleration of the puck is approximately 421.47 m/s^2.
To find the instantaneous acceleration of the puck, we need to apply Newton's second law of motion, which states that the acceleration of an object is equal to the net force acting on it divided by its mass.
In this case, we have two forces acting on the puck simultaneously, one directed west and the other at a 30 degree angle east of north. To find the net force, we need to determine the horizontal and vertical components of the two forces.
1.30 * 10^3 N west can be decomposed into two components: one in the horizontal (x) direction and one in the vertical (y) direction. Since it is directed west, the x-component will be negative.
Horizontal component: Fx = -1.30 * 10^3 N
Vertical component: Fy = 0 N (no vertical component in this force)
9.90 * 10^2 N at 30 degrees east of north can also be decomposed into two components. The angle is measured east of north, so the y-component will be positive.
Horizontal component: Fx = 9.90 * 10^2 N * cos(30°)
Vertical component: Fy = 9.90 * 10^2 N * sin(30°)
Now, we can find the net force by adding the horizontal and vertical components.
Fx_net = -1.30 * 10^3 N + 9.90 * 10^2 N * cos(30°)
Fy_net = 9.90 * 10^2 N * sin(30°)
Next, we can calculate the net force magnitude using the Pythagorean theorem:
F_net = √(Fx_net^2 + Fy_net^2)
Finally, we can substitute the net force and the given mass into Newton's second law to find the instantaneous acceleration:
acceleration = F_net / mass
Substitute the values and solve the equation to find the instantaneous acceleration of the puck.
1.30*10^3 @ W = (-1300,0)
9.90*10^2 @ N30E = (445,857)
add the forces to get (-855,857) = 1211 @ N45W
Since F=ma, a = 1211/.159 = 7613 m/s^2 in that direction