1. **Problem statement:** We have a particle of mass $m=2000$ kg moving along the $x$-axis from $x=0$ to $x=9.0$ m under a force $F_a$. The acceleration $a(x)$ vs. position $x$ graph is given with scale $a_s=6.0$ m/s². We need to find: - (a) The work done by the force at $x=4.0$ m, $7.0$ m, and $9.0$ m. - (b) The velocity magnitude and direction at these points. 2. **Relevant formulas and concepts:** - Work done by force: $W = \int F \, dx = m \int a(x) \, dx$ since $F=ma$. - Velocity from work-energy theorem: $W = \Delta K = \frac{1}{2}m v^2 - \frac{1}{2}m v_0^2$. - Initial velocity $v_0=0$ (starts from rest). 3. **Acceleration function from graph:** - From $x=0$ to $x=2$: $a(x)$ increases linearly from 0 to $a_s=6$ m/s². So, $a(x) = 3x$ (since slope $= \frac{6-0}{2-0}=3$). - From $x=2$ to $x=4$: $a(x) = 6$ m/s² (constant). - From $x=4$ to $x=6$: $a(x)$ decreases linearly from $6$ to $-6$ m/s². Slope $= \frac{-6-6}{6-4} = -6$ m/s² per m. Equation: $a(x) = 6 - 6(x-4) = 30 - 6x$. - From $x=6$ to $x=8$: $a(x) = -6$ m/s² (constant). - From $x=8$ to $x=9$: $a(x)$ increases linearly from $-6$ to $0$. Slope $= \frac{0 - (-6)}{9-8} = 6$. Equation: $a(x) = -6 + 6(x-8) = 6x - 54$. 4. **Calculate work done $W(x) = m \int_0^x a(x) dx$ at $x=4,7,9$ m:** - For $x=4$ m: \[ \int_0^4 a(x) dx = \int_0^2 3x dx + \int_2^4 6 dx = \left[ \frac{3x^2}{2} \right]_0^2 + [6x]_2^4 = \frac{3 \times 4}{2} + (24 - 12) = 6 + 12 = 18 \] So, $W(4) = 2000 \times 18 = 36000$ J. - For $x=7$ m: \[ \int_0^7 a(x) dx = \int_0^2 3x dx + \int_2^4 6 dx + \int_4^6 (30 - 6x) dx + \int_6^7 (-6) dx \] Calculate each: - $\int_0^2 3x dx = 6$ - $\int_2^4 6 dx = 12$ - $\int_4^6 (30 - 6x) dx = [30x - 3x^2]_4^6 = (180 - 108) - (120 - 48) = 72 - 72 = 0$ - $\int_6^7 (-6) dx = -6$ Sum: $6 + 12 + 0 - 6 = 12$ So, $W(7) = 2000 \times 12 = 24000$ J. - For $x=9$ m: \[ \int_0^9 a(x) dx = \int_0^2 3x dx + \int_2^4 6 dx + \int_4^6 (30 - 6x) dx + \int_6^8 (-6) dx + \int_8^9 (6x - 54) dx \] Calculate each: - $\int_0^2 3x dx = 6$ - $\int_2^4 6 dx = 12$ - $\int_4^6 (30 - 6x) dx = 0$ (from above) - $\int_6^8 (-6) dx = -12$ - $\int_8^9 (6x - 54) dx = [3x^2 - 54x]_8^9 = (243 - 486) - (192 - 432) = -243 - (-240) = -3$ Sum: $6 + 12 + 0 - 12 - 3 = 3$ So, $W(9) = 2000 \times 3 = 6000$ J. 5. **Calculate velocity at these points using $W = \frac{1}{2} m v^2$ (since $v_0=0$):** - At $x=4$ m: \[ 36000 = \frac{1}{2} \times 2000 \times v^2 \Rightarrow v^2 = \frac{36000 \times 2}{2000} = 36 \Rightarrow v = 6 \text{ m/s} \] - At $x=7$ m: \[ 24000 = 1000 v^2 \Rightarrow v^2 = 24 \Rightarrow v = \sqrt{24} \approx 4.90 \text{ m/s} \] - At $x=9$ m: \[ 6000 = 1000 v^2 \Rightarrow v^2 = 6 \Rightarrow v = \sqrt{6} \approx 2.45 \text{ m/s} \] 6. **Direction of velocity:** - Velocity is positive (forward) if work done is positive and acceleration is mostly positive. - Since acceleration becomes negative after $x=6$, velocity decreases but remains positive (particle still moves forward). **Final answers:** - Work done by force: - At $x=4.0$ m: $36000$ J - At $x=7.0$ m: $24000$ J - At $x=9.0$ m: $6000$ J - Velocity magnitude and direction: - At $x=4.0$ m: $6$ m/s forward - At $x=7.0$ m: $4.90$ m/s forward - At $x=9.0$ m: $2.45$ m/s forward