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The answer is B apparently. The way I approached this problem was to draw a free body diagram. While on the incline $$mgcos\theta$$ cancels with the normal force. So that leaves $$mgsin\theta$$ as the force that causes the acceleration while on the incline.

Work = $$Fscos\phi$$ So on the incline W = $$(mgsin\theta)(s)(1) = (mgsin\theta)(s) $$

While on the second part of the motion the only force causing the deceleration is the friction force. so $$W_2 = (-F_f)(x)(-1)$$

$$W_2 = (\mu_k)(mg)(x)$$

I just don't see where I messed up or what I am misunderstanding because apparently $W_{tot} = 0$

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You are missing a negative sign.

Consider the block as the system.

You could now use the work-energy theorem which states that the change in the kinetic energy of the block is equal to the work done on the block.

The block starts at rest and finishes at rest so the change in kinetic energy of the block is zero.
Hence the work done on the block is zero.

The work done on the block by gravitational force (external force) is $mg\sin\theta\,s$.

The work done on the block by the frictional force (external force) is $-\mu_{\rm k}mg\,x$.
The negative sign is there because the frictional force is to the left and the displacement of the force is to the right.

Total work done on the block by the external forces is $mg\sin\theta\,s-\mu_{\rm k}mg\,x$ and this is zero because there is no change in the kinetic energy of the block.

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  • $\begingroup$ Okay this makes sense, since the Work energy theorem states work is = to change in kinetic energy so if an object starts at rest and ends at rest the work is = to 0. I noticed you said that I missed a sign, isn't friction negative because it is to the left, and when you calculate work friction does on the block shouldn't that cancel the negative? Unless the negative you put in your example is from $-g$? $\endgroup$
    – Jude
    Mar 25, 2017 at 19:22
  • $\begingroup$ The negative is because the force is to the left and the displacement is to the right. $\endgroup$
    – Farcher
    Mar 25, 2017 at 22:28
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The block does not gain KE. If the blocks KE does not change, the total work done has external forces must sum to zero, otherwise the block would end up with more KE than it started off with.

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  • $\begingroup$ So I cannot show that the total work will be 0?, Through algebra? Instead I need to know that if energy is conserved then the block at the end has the same amount energy it started with? $\endgroup$
    – Jude
    Mar 25, 2017 at 6:43
  • $\begingroup$ Yes, unless you are told that the initial and final velocities are not the same and you have some way of determining the relationship between s, x, and uk. $\endgroup$
    – H. Chandrasekaran
    Mar 25, 2017 at 6:46
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To see that the net work done on the block is indeed zero through explicit algebra, one can do some simple kinematics.

WLOG, let $t=0$ be where the block is at the bottom of the ramp. Then its speed here is $v_b(t=0) = v_b = \sqrt{2gs \sin \theta}$. Thereafter, the block is acted upon by a kinetic friction force such that $$m \ddot{x} = -f = - mg \mu_k \Rightarrow \dot{x} = -g\mu_k t + v_b\,\,\,\text{and}\,\,\,x = -g\mu_k \frac{t^2}{2} + v_b t$$ with $x(t=0) = x_0 = 0$. The block thus comes to rest in a time $$T = \frac{v_b}{g \mu_k}$$ and has travelled a distance $$x_{\text{max}} = \frac{s \sin \theta}{\mu_k}\,\,\,\,\,\,\,\,\,\,(1)$$

Now, as noted in the answer by Farcher, the total work done on the block by the external forces throughout the whole motion is $$W_{\text{tot}} = m g \sin \theta \, s - \mu_k\, mg \, x_{\text{max}}\,\,\,\,\,\,\,\,(2).$$ If we multiply $(1)$ by $mg$ and rearrange we obtain $$m g \sin \theta \, s - \mu_k\, mg \, x_{\text{max}} = 0$$ which by $(2)$ is equal to $W_{\text{tot}}$. Hence, $$W_{\text{tot}} = 0.$$

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The work exerted on the block because the gravity is equal to kinetic energy loss because the friction (this using energy conservation principle). If the final velocity of the block is zero, the total work is zero too, other wise the block continues moving.

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Net work equals change in energy. To figure out if work was done, calculate the initial energy of the block and compare it to the final energy of the block.

Initial energy: The block just has Potential Energy, and the formula for Ep = mgh. The formula for this would be (m)(9.8)(s * sin_theta)

Final energy: The block just has Kinetic energy, and the formula for Ek = 0.5mv^2. Here you need to figure out the box's velocity when it finishes sliding down the ramp and before it starts being slowed down by friction on the ground. Since there was no friction on the ramp, the mgh of the box at the top of the ramp = the 0.5mv^2 of the box at the bottom of the ramp, due to the Conservation of Energy (since there was no friction on the ramp).

So, we know that the box's energy at the top of the ramp and right at the bottom are equal. But since the box experiences a frictional force when it's on the ground, it loses some of its kinetic energy due to friction. When it stops, its final kinetic energy is less than its potential energy at the top of the ramp, since we know its kinetic energy at the bottom of the ramp was equal to the Ep at the top.

So work was done by friction on the ground - your textbook could have been wrong with the answer it gave.

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