# The work due to friction given an influential force

Statement: I am trying to work through Chapter 8, Problem #70 of OpenStax University Physics Volume 1 and I cannot figure out what I'm doing wrong. The problem statement is as follows:

A particle of mass $$2.0 \, \rm{kg}$$ moves under the influence of the force $$F(x) = \left(-5x^2+7x\right) \, \rm{N}$$. Suppose a frictional force also acts on the particle. If the particle's speed when it starts at $$x = -4.0 \, \rm{m}$$ is $$0.0 \, \rm{m/s}$$ and when it arrives at $$x = 4.0 \, \rm{m}$$ is $$9.0 \, \rm{m/s}$$, how much work is done on it by the frictional force between $$x = -4.0 \, \rm{m}$$ and $$x = 4.0 \, \rm{m}$$?

My attempt: Using conservation of energy and taking into account the dissipating frictional force, the formula is:

$$W_f = \Delta K + \Delta U.$$

The change in kinetic energy is computable because both the initial and final velocities are given:

$$\Delta K = \dfrac{1}{2}m\left(v_f^2-v_i^2\right) = 0.5(2 \, \rm{kg})\big((9.0 \, \rm{m/s})^2 - (0.0 \, \rm{m/s})^2\big) = 81 \, \rm{J}.$$

The change in potential energy is computed via the integral:

$$\Delta U = -\int_{x_0}^{x_f} F(x) \, dx = -\int_{-4}^4 \left(-5x^2+7x\right) \, dx = 213.33 \, \rm{J}$$

so altogether I get that $$W_f = 294.33 \, \rm{J}$$ which cannot be correct. The solution manual I have lists $$130 \, \rm{J}$$, which also doesn't make sense since the work due to frictional force cannot be positive.

Question: What am I misunderstanding? I realize that if I subtract my $$\Delta K$$ from my $$\Delta U$$, I get their answer, but I cannot for the life of me understand why that would be correct. Thank you in advance!

• The question seems problematic. If a particle starts at $x=-4$ at rest, it would never reach $x=4$ under the given forces. The force $F=-5x^2+7x$ always points towards the negative $x$ direction for $x<0$. – Feynmans Out for Grumpy Cat May 18 at 8:31
• If the force, in fact, has a negative overall sign then such a motion would be possible and then the sign of your $\Delta U$ would also get reversed and then you would get $W_f\approx-130$ J. This would still be not in agreement with your manual tho due to the sign. :/ – Feynmans Out for Grumpy Cat May 18 at 8:35
• What is the nature of the potential energy? – Bob D May 18 at 9:00
• @DvijMankad Yeah, I'm just not sure. Maybe they meant to have the velocities swapped? – Suugaku May 18 at 21:28

What you are calling $$\Delta U$$ is simply the work done by the force $$f(x)$$ on the particle. If there were no friction, by the work-energy theorem that work done (213 j) on the particle would equal its change (increase) in kinetic energy. Since the actual increase was only 81 J, friction had to do negativie work of about 130 J. The friction work has to be negative because it transfers energy from the particle.
• I am still a bit confused. The work-energy theorem says $W_{net} = \Delta K \implies W_{F} + W_{fric} = \Delta K \implies -213 \, \rm{J} + W_{fric} = 81 \, \rm{J} \implies W_{fric} = 294 \, \rm{J}$, so the same as I derived above? Note that $\Delta U$ above is potential energy so $\Delta U = -W_{F}$. – Suugaku May 18 at 21:29
• I apologize, I must have a misunderstanding of potential energy. I thought in its most general form that potential energy is defined to be opposite of the work done. So the nature is determined by the force? In any case, we don't need to worry about it and can consider the work-energy theorem as you suggested. However, the issue is the signs. The work done by the force is actually negative: $-213 \, \rm{J}$, and subtracting this from the change in kinetic energy gives positive work due to friction, which is impossible. – Suugaku May 19 at 21:36