1. **State the problem:** We have a block of mass $m=7$ kg moving at initial speed $v=2$ m/s on a flat surface with kinetic friction coefficient $\mu_k=0.19$. We need to find: (a) Initial kinetic energy (b) Magnitude of net force on the block (c) Magnitude of acceleration after the push (d) Distance traveled before stopping (e) Net work done on the block until it stops (f) Change in kinetic energy from push to stop 2. **Calculate initial kinetic energy:** $$KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 7 \times 2^2 = \frac{1}{2} \times 7 \times 4 = 14\,J$$ 3. **Calculate net force magnitude:** The only horizontal force after the push is kinetic friction: $$F_{fric} = \mu_k m g = 0.19 \times 7 \times 9.8 = 13.054\,N$$ The net force magnitude is $13.054$ N (opposes motion). 4. **Calculate acceleration magnitude:** Using Newton's second law: $$a = \frac{F_{net}}{m} = \frac{13.054}{7} = 1.865\,m/s^2$$ 5. **Calculate stopping distance:** Using kinematic equation with final velocity $v_f=0$: $$v_f^2 = v^2 - 2 a d \implies 0 = 2^2 - 2 \times 1.865 \times d$$ Solve for $d$: $$d = \frac{4}{2 \times 1.865} = \frac{4}{3.73} = 1.073\,m$$ 6. **Calculate net work done:** Work done by friction (net work) equals change in kinetic energy: $$W = F_{fric} \times d \times \cos 180^\circ = -13.054 \times 1.073 = -14\,J$$ 7. **Calculate change in kinetic energy:** $$\Delta KE = KE_{final} - KE_{initial} = 0 - 14 = -14\,J$$ **Final answers:** (a) 14 (b) 13.054 (c) 1.865 (d) 1.073 (e) -14 (f) -14