Two constant forces act on an object of mass m = 4.50 kg object moving in the xy plane as shown in the figure below. Force 1 is 21.0 N at 35.0°, and force 2 is 47.0 N at 150°. At time t = 0, the object is at the origin and has velocity (3.70 + 2.25) m/s.

(a) Express the two forces in unit-vector notation.
1 = N
2 = N

(b) Find the total force exerted on the object.
N

(c) Find the object's acceleration.
m/s2

Now, consider the instant t = 3.00 s.
(d) Find the object's velocity.
m/s

(e) Find its position.
m

(f) Find its kinetic energy from ½mvf2.
kJ

(g) Find its kinetic energy from ½mvi2 + Σ · Δ.
kJ

(h) What conclusion can you draw by comparing the answers to parts (f) and (g)?

To solve this problem, we'll break it down into several steps:

Step 1: Expressing the forces in unit-vector notation
To express the forces in unit-vector notation, we'll use the trigonometric definitions of the unit vectors in the x and y directions. The x-direction unit vector is î and the y-direction unit vector is ĵ.

Force 1 is given as 21.0 N at 35.0°. To express it in unit-vector notation, we'll first find its components in the x and y directions using trigonometry. The x-component of Force 1 is given by F1x = F1 * cos(θ), where θ is the angle (35.0°) and F1 is the magnitude (21.0 N). The y-component of Force 1 is given by F1y = F1 * sin(θ). Substitute the values to find F1x and F1y.

Force 2 is given as 47.0 N at 150°. Similarly, we'll find its components in the x and y directions using trigonometry. The x-component of Force 2 is given by F2x = F2 * cos(θ), where θ is the angle (150°) and F2 is the magnitude (47.0 N). The y-component of Force 2 is given by F2y = F2 * sin(θ). Substitute the values to find F2x and F2y.

Step 2: Finding the total force exerted on the object
The total force exerted on the object is the vector sum of Force 1 and Force 2. To find the total force, add the components of Force 1 and Force 2 in the x and y directions separately. The x-component of the total force is Fx = F1x + F2x, and the y-component of the total force is Fy = F1y + F2y. Express the total force in unit-vector notation as F = Fx * î + Fy * ĵ.

Step 3: Finding the object's acceleration
The object's acceleration is given by Newton's second law of motion, F = ma, where F is the total force exerted on the object and m is the mass of the object. Rearrange the equation to solve for acceleration, a = F / m. Substitute the values to find the acceleration.

Now, consider the instant t = 3.00 s.

Step 4: Finding the object's velocity
To find the object's velocity at t = 3.00 s, we'll use the equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Substitute the given initial velocity and acceleration to find the final velocity.

Step 5: Finding the object's position
To find the object's position at t = 3.00 s, we'll use the equation of motion, s = ut + 0.5at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. Substitute the given initial velocity, acceleration, and time to find the displacement.

Step 6: Finding the object's kinetic energy from ½mvf^2
To find the object's kinetic energy at t = 3.00 s, we'll use the equation for kinetic energy, KE = 0.5 * m * vf^2, where KE is the kinetic energy, m is the mass, and vf is the final velocity. Substitute the given mass and final velocity to find the kinetic energy.

Step 7: Finding the object's kinetic energy from ½mvi^2 + Σ · Δ
To find the object's kinetic energy at t = 3.00 s using the work-energy theorem, we'll use the equation KE = 0.5 * m * vi^2 + Σ · Δ, where KE is the kinetic energy, m is the mass, vi is the initial velocity, Σ corresponds to the dot product of force and displacement, and Δ is the displacement. Calculate the dot product Σ · Δ using the components of the total force and displacement. Substitute the given mass, initial velocity, and displacement to find the kinetic energy.

Step 8: Comparing the answers to parts (f) and (g)
Compare the values obtained for kinetic energy from parts (f) and (g). If the values are equal or very close, it indicates the conservation of mechanical energy. If they are different, there is some non-conservative force at work, causing a change in the object's mechanical energy.