Problem 1- A 4 kg puck can slide with negligible friction over a horizontal surface, taken as the xy plane.

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The puck has a velocity of 3i m / s at one instant ti = 0 . Eight seconds later, its velocity is to be
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8i + 10 j m / s . Find the magnitude and direction of the net force acting on the puck.

the later speed is 10.

So, the acceleration is 10m/s /8s = 1.25 m/s^2

F = ma 3*1.25 = 5N

To find the magnitude and direction of the net force acting on the puck, we need to apply Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. The acceleration can be calculated by taking the difference in velocity and dividing it by the time interval.

Given:
Mass of the puck, m = 4 kg
Velocity at ti=0, v1 = 3i m / s
Velocity at ti=8s, v2 = 8i + 10j m / s
Time interval, Δt = 8 seconds

Step 1: Calculate the acceleration
Acceleration, a = (v2 - v1) / Δt
= (8i + 10j - 3i) / 8
= (5i + 10j) / 8
= 5/8 i + 10/8 j
= (5/8)i + (5/4)j m/s^2

Step 2: Calculate the net force
Net force, F = m * a
= 4kg * ((5/8)i + (5/4)j) m/s^2
= (20/8)i + (20/4)j
= (5/2)i + 5j N

Step 3: Calculate the magnitude and direction of the net force
Magnitude of the net force, |F| = √[(5/2)^2 + 5^2]
= √[25/4 + 25]
= √(25/4 + 100/4)
= √(125/4)
= √125/2
= (5√5)/2 N

Direction of the net force, θ = tan^(-1)(Fy/Fx)
= tan^(-1)(5/(5/2))
= tan^(-1)(10/5)
= tan^(-1)(2)
= 63.43° (approx)

Therefore, the magnitude of the net force acting on the puck is (5√5)/2 N, and its direction is approximately 63.43° (counterclockwise from the positive x-axis).

To find the magnitude and direction of the net force acting on the puck, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

In this case, we are given the initial and final velocities of the puck, and we need to find the net force acting on it. We can start by finding the acceleration of the puck using the equation:

acceleration = (change in velocity) / (time)

The change in velocity can be calculated by subtracting the initial velocity from the final velocity:

change in velocity = final velocity - initial velocity

= (8i + 10j) m/s - (3i) m/s

= (8i + 10j) m/s - (3i - 0j) m/s

= (8i + 10j - 3i + 0j) m/s

= (5i + 10j) m/s

Now, we can substitute the values into the equation for acceleration:

acceleration = (5i + 10j) m/s / 8 s

= (5/8)i + (10/8)j m/s^2

= (5/8)i + (5/4)j m/s^2

Next, we can find the magnitude of the acceleration using the equation:

magnitude of acceleration = sqrt((x-component of acceleration)^2 + (y-component of acceleration)^2)

magnitude of acceleration = sqrt((5/8)^2 + (5/4)^2) m/s^2

= sqrt(25/64 + 25/16) m/s^2

= sqrt(25/64 + 100/64) m/s^2

= sqrt(125/64) m/s^2

= (5/8)sqrt(5) m/s^2

Finally, we can calculate the net force using Newton's second law:

net force = mass * acceleration

Here, the mass of the puck is given as 4 kg:

net force = 4 kg * ((5/8)sqrt(5) m/s^2)

= (20/8)sqrt(5) N

= (5/2)sqrt(5) N

So, the magnitude of the net force acting on the puck is (5/2)sqrt(5) N.

To find the direction of the net force, we can use the direction of the acceleration, which in this case is given by the components (5/8)i + (5/4)j m/s^2.

Therefore, the direction of the net force is in the same direction as the acceleration vector, which is at an angle of arctan((5/4)/(5/8)) = arctan(2) with the positive x-axis.

Hence, the direction of the net force acting on the puck is arctan(2) from the positive x-axis.