# A force acts on a particle based on position, determine work. Sign question.

 P: 5 I have a question about this problem in relation to the way the solution manual handles it. 1. The problem statement, all variables and given/known data A force Fx acts on a particle of mass 1.5kg. The force is related to the position x of the particle by the formula Fx = Cx3, where C = .50. IF x is in maters and Fx is in newtons. (a) What are the SI units of C? (b) Find the work done by this force as the particle moves from x = 3.0m to x = 1.5 m. (c) At x = 3.0m the force points in the opposite direction of the particles velocity (speed is 12m/s) What is its speed at x = 1.5m? Ignore (a) 2. Relevant equations $W = \int F_{x} dx$ $W = \frac{1}{2}{mv}^{2}_{f} - \frac{1}{2}{mv}^{2}_{i}$ Fx = Cx3 3. The attempt at a solution Okay, when I solved the problem I used the first equation which gave me -9.5 Joules. $W = {\int}^{1.5m}_{3.0m} .5x^{3} dx = -9.5 Joules$ However, when I referenced the solution manual, they placed a negative sign in front of Fx in the equation because the force is "in the opposite direction of the displacement." $W = {\int}^{1.5m}_{3.0m} -.5x^{3} dx = 9.5 Joules$ This made no sense to me because the velocity of the particle is, from what I can tell, equal to -12m/s because the force function is always positive. So, it only makes sense to me that the force should be slowing the particle down since the particle is moving in the opposite direction of the force. Consequently, because of the sign conversion, this throws off the answer to c as well because when I calculated the velocity with the negative work the velocity came out to 11.46m/s however, because the manual claims the work was positive, their result was 13m/s. Derived Equation: $\sqrt{\frac{2W+m{v}^{2}_{i}}{m}} = v_f$ I can not make sense of why the solution manual flipped the sign of the force function.
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,683 The solution manual is wrong. You're right. What the solution manual is trying to explain is that the force ##\vec{F}## and the displacement ##d\vec{x}## point in the opposite direction, so $$dW=\vec{F}\cdot d\vec{x} = |\vec{F}||d\vec{x}|\cos 180^\circ = -|\vec{F}||d\vec{x}|$$What the person who wrote the solution missed was the fact that ##|d\vec{x}| = -dx## because the lower limit of the integral is greater than the upper limit, so the work is given by $$W = -\int |\vec{F}||d\vec{x}| = +\int_{3.0\text{ m}}^{1.5\text{ m}} Cx^3\,dx$$ as you had.