Two hockey players strike a puck of mass 0.168 kg with their sticks simultaneously, exerting forces of 1.16 103 N, directed west, and 9.90 102 N, directed 30.0° east of north. Find the instantaneous acceleration of the puck.
Well, this is quite a situation on the ice! But fear not, I'm here to help you calculate the acceleration of that speedy puck. Let's break it down step by step.
First, we need to determine the net force acting on the puck. To do that, we'll resolve the forces into their respective x and y components.
The force exerted by the first player in the west direction is equal to 1.16 x 10^3 N, and since it's directed solely west, the x-component of this force will be 1.16 x 10^3 N.
Now, for the force exerted by the second player, which is directed 30.0° east of north. We need to find the x and y components of this force, so buckle up!
The angle between the force and the north direction is 30.0°, so the y-component of this force would be (9.90 x 10^2 N) * sin(30.0°).
The x-component of this force would be (9.90 x 10^2 N) * cos(30.0°).
So now we have the x and y components of both forces. Adding up all the x components gives us the net force in the x direction, and adding up the y components gives us the net force in the y direction.
Lastly, we'll use Newton's second law of motion, F = ma. Since we now have the net force, we can divide it by the mass of the puck (0.168 kg) to calculate the instantaneous acceleration. Hooray for physics!
I hope this explanation didn't leave you feeling "puck"ish! Just remember to double-check your math and use the proper unit conversions. Let me know if you need any further assistance!
To find the instantaneous acceleration of the puck, we can use 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.
1. First, we need to find the net force acting on the puck. Since the two players are striking the puck simultaneously, the forces they exert can be added together to find the net force.
2. The force exerted by the first player is 1.16 * 10^3 N, directed west. Since the force is already given in the desired direction, we don't need to break it down into its x and y components.
3. The force exerted by the second player is 9.90 * 10^2 N, directed 30.0° east of north. We need to resolve this force into its x and y components to find the net force in the x and y direction.
To find the x component, we can use the cosine function:
Fx = F * cosθ
Fx = (9.90 * 10^2 N) * cos(30°)
Fx = (9.90 * 10^2 N) * √3/2
To find the y component, we can use the sine function:
Fy = F * sinθ
Fy = (9.90 * 10^2 N) * sin(30°)
Fy = (9.90 * 10^2 N) * 1/2
4. Now we can find the net force by adding the x and y components. Since the x component is directed east and the y component is directed north, we subtract the x component and add the y component.
F_net_x = - (9.90 * 10^2 N) * √3/2
F_net_y = (9.90 * 10^2 N) * 1/2
F_net = (F_net_x)î + (F_net_y)ĵ
F_net = - (9.90 * 10^2 N) * √3/2 î + (9.90 * 10^2 N) * 1/2 ĵ
5. Now we can find the net force acting on the puck using the Pythagorean theorem:
|F_net| = √(F_net_x)^2 + (F_net_y)^2
6. Substitute the values into the equation to find the magnitude of the net force acting on the puck.
7. Finally, to find the instantaneous acceleration of the puck, divide the net force by the mass of the puck.
|a| = |F_net| / m
8. Substitute the values to calculate the instantaneous acceleration of the puck.
30o E. of N. = 60o CCW.
Fr = -1160 + 990[60].
Fr = -1160 + 990*Cos60+990*sin60,
Fr = -1160 + 495+857.4i,
Fr = -665 + 857.4i = 1085N[-52.2] = 1085N[127.8o] = Resultant force.
Fr = M*a = 1085[127.8o].
0.168a = 1085[127.8,
a = 6458 m/s^2[127.8o].