4. Determine the velocity of a 2kg donut sitting on a horizontal surface with a coefficient of friction 0.1 after 2 seconds, with the following forces acting on it. 5N[S],15N[S33W],6N[E] and 16N[N27E

Here's how to determine the donut's velocity:1. **Resolve forces into components:** We need to break down the forces into their x (horizontal) and y (vertical) components.* 5N[S]: x = 0 N, y = -5 N* 15N[S33W]: x = -15 * sin(33°) ≈ -8.17 N, y = -15 * cos(33°) ≈ -12.58 N* 6N[E]: x = 6 N, y = 0 N* 16N[N27E]: x = 16 * sin(27°) ≈ 7.26 N, y = 16 * cos(27°) ≈ 14.26 N2. **Calculate net forces:** Sum the x and y components separately.* Net Force in x (F_x) = 0 N - 8.17 N + 6 N + 7.26 N ≈ 5.09 N* Net Force in y (F_y) = -5 N - 12.58 N + 0 N + 14.26 N ≈ -3.32 N3. **Determine the normal force:** The normal force (F_N) counteracts the net vertical force. Since the donut is on a horizontal surface, the normal force will be equal in magnitude but opposite in direction to the net vertical force plus the force due to gravity.* Force due to gravity (F_g) = m * g = 2 kg * 9.8 m/s² ≈ 19.6 N (downwards, so negative)* F_N = - (F_y + F_g) = - (-3.32 N - 19.6 N) ≈ 22.92 N4. **Calculate the force of friction:** Friction opposes motion and is calculated using:* Force of friction (F_f) = μ * F_N = 0.1 * 22.92 N ≈ 2.29 N5. **Calculate net horizontal force (accounting for friction):** Since friction opposes motion, we subtract it from the net force in the x-direction.* Net Force in x (with friction) = F_x - F_f = 5.09 N - 2.29 N ≈ 2.8 N6. **Calculate acceleration:** Use Newton's second law (F = ma)* Acceleration (a) = Net Force / mass = 2.8 N / 2 kg ≈ 1.4 m/s²7. **Calculate final velocity:** Assuming the donut starts from rest (initial velocity = 0 m/s), we can use the following kinematic equation:* Final velocity (v) = Initial velocity (u) + acceleration (a) * time (t)* v = 0 m/s + 1.4 m/s² * 2 s = 2.8 m/sTherefore, the velocity of the donut after 2 seconds is approximately in a direction slightly North of East.