Solving Newton's Second Law: Tension, Work and Kinetic Energy

[greenrichy]

Homework Statement

A person drags a $75 kg$ block using a rope over a rough surface that is inclined $\theta=41^{\circ}$ above the horizontal (the block's initial velocity is zero). The block is dragged a distance of $30.5 m$ along the incline with a constant acceleration of $0.25\frac{m}{s^2}$. The coefficient of kinetic friction between the block and the ground is $0.31$.

(a) Find the total work done by the person who is pulling the block (you may assume that the rope is parallel with the surface during the drag)

(b) The instant before the block reaches the top of the hill, it snaps. The block slides back down the hill. Determine the block's velocity the instant it reaches the bottom of the hill.

Relevant Equations

$$\sum F_x = T - w_x - f_k = ma_x $$

$$\sum F_x = T - w_x - f_k = ma_x $$
$$ T = mg\sin(\theta) + mg\cos(\theta)\mu_k + ma_x$$
$$ T = (9.8 \frac{m}{s^2}) \cdot (\sin(41^{\circ}) + \cos(41^{\circ})) + (75kg)\cdot(0.25\frac{m}{s^2}) $$
$$T = 672.91 N $$

Having found the tension force, I can find the work done by the person who's pulling the block:
$$ W = 672.91N \cdot 30.5m $$
$$ W = 20510.2968J $$

Quick question: can I solve the above part with the Work-Energy theorem?

Now, when it comes to part (b), I've solved it using the Work-Energy theorem:

$$ W = \Delta K $$
$$ (-f_k - w_x)\cdot \cos(0^{\circ}) \cdot (-d) = \frac{1}{2} m\nu^2$$
$$ \nu=\sqrt{2g(\cos(\theta)\mu_k + \sin(\theta))\cdot d} $$

After substituting values, I got the following answer for the block's final speed:

$$ \nu = 23.0587179 \frac{m}{s^2} $$

But how do I express the object's final velocity? Would the following suffice?

The object travels with a speed of $23.1 \frac{m}{s^2}$ at an angle of $41^{\circ}$.

 

[collinsmark]

Hello @greenrichy,

Welcome to PF!

Problem Statement: A person drags a $75 kg$ block using a rope over a rough surface that is inclined $\theta=41^{\circ}$ above the horizontal (the block's initial velocity is zero). The block is dragged a distance of $30.5 m$ along the incline with a constant acceleration of $0.25\frac{m}{s^2}$. The coefficient of kinetic friction between the block and the ground is $0.31$.

(a) Find the total work done by the person who is pulling the block (you may assume that the rope is parallel with the surface during the drag)

(b) The instant before the block reaches the top of the hill, it snaps. The block slides back down the hill. Determine the block's velocity the instant it reaches the bottom of the hill.
Relevant Equations: $$\sum F_x = T - w_x - f_k = ma_x $$

$$\sum F_x = T - w_x - f_k = ma_x $$
$$ T = mg\sin(\theta) + mg\cos(\theta)\mu_k + ma_x$$
$$ T = (9.8 \frac{m}{s^2}) \cdot (\sin(41^{\circ}) + \cos(41^{\circ})) + (75kg)\cdot(0.25\frac{m}{s^2}) $$
$$T = 672.91 N $$

Having found the tension force, I can find the work done by the person who's pulling the block:
$$ W = 672.91N \cdot 30.5m $$
$$ W = 20510.2968J $$

Quick question: can I solve the above part with the Work-Energy theorem?

Now, when it comes to part (b), I've solved it using the Work-Energy theorem:

$$ W = \Delta K $$
$$ (-f_k - w_x)\cdot \cos(0^{\circ}) \cdot (-d) = \frac{1}{2} m\nu^2$$
$$ \nu=\sqrt{2g(\cos(\theta)\mu_k + \sin(\theta))\cdot d} $$

After substituting values, I got the following answer for the block's final speed:

$$ \nu = 23.0587179 \frac{m}{s^2} $$

But how do I express the object's final velocity? Would the following suffice?

The object travels with a speed of $23.1 \frac{m}{s^2}$ at an angle of $41^{\circ}$.

Aside from minor rounding errors and/or issues with significant digits, your answer for part (a) looks correct to me.

Part (b) of the problem statement is worded a bit wonky. I'm honestly not sure how to interpret it. But I think it's one of two choices:

1. After pulling the block 30.5 meters up the incline at a constant acceleration, the block "magically" and instantly decelerates to a standstill before sliding back down the incline.
2. After pulling the block 30.5 meters up the incline at a constant acceleration, the person releases the rope allowing the block to slide up to the very top of the hill under its own inertia, at which point the rope breaks (which is a bit odd since the rope is not under tension at this point), and then the block slides back down the incline (note the total length of the incline is a bit longer than 30.5 meters).

I think you might have gone with interpretation 1., but I would have gone with interpretation 2., since it seems slightly more reasonable to me.

I worked out interpretation 2., and the answer is a bit different than what you had.

 

[greenrichy]

Hello @greenrichy,

Welcome to PF!
Aside from minor rounding errors and/or issues with significant digits, your answer for part (a) looks correct to me.

Part (b) of the problem statement is worded a bit wonky. I'm honestly not sure how to interpret it. But I think it's one of two choices:

1. After pulling the block 30.5 meters up the incline at a constant acceleration, the block "magically" and instantly decelerates to a standstill before sliding back down the incline.
2. After pulling the block 30.5 meters up the incline at a constant acceleration, the person releases the rope allowing the block to slide up to the very top of the hill under its own inertia, at which point the rope breaks (which is a bit odd since the rope is not under tension at this point), and then the block slides back down the incline (note the total length of the incline is a bit longer than 30.5 meters).

I think you might have gone with interpretation 1., but I would have gone with interpretation 2., since it seems slightly more reasonable to me.

I worked out interpretation 2., and the answer is a bit different than what you had.

Thank you for your reply!

Sorry about the confusion in part (b). I interpreted it as follows:

After pulling the block 30.5 meters up the incline at a constant acceleration, the person stops (the block's velocity becomes zero) and releases the rope causing the block slide down the hill.

 

[collinsmark]

Thank you for your reply!

Sorry about the confusion in part (b). I interpreted it as follows:

After pulling the block 30.5 meters up the incline at a constant acceleration, the person stops (the block's velocity becomes zero) and releases the rope causing the block slide down the hill.

Yes, that seems to be your interpretation from your solution.

I would have gone with the second interpretation: Just because the person stops pulling doesn't mean the block stops moving. Newton's first law of motion and all.

[Edit: if you have the chance, I would ask your instructor for clarification.]

 

[haruspex]

The object travels with a speed of $23.1 \frac{m}{s^2}$

Check the units.

After pulling the block 30.5 meters up the incline at a constant acceleration, the person releases the rope allowing the block to slide up to the very top of the hill under its own inertia, at which point the rope breaks

More reasonably, after 30.5m the rope breaks but the object continues up a little before sliding down. Same answer, though.

 

[neilparker62]

$$ T = (9.8 \frac{m}{s^2}) \cdot (\sin(41^{\circ}) + \cos(41^{\circ})) + (75kg)\cdot(0.25\frac{m}{s^2}) $$

Some numbers left out here but you must have included them when you did the calculation.

 

[PeroK]

Problem Statement: A person drags a $75 kg$ block using a rope over a rough surface that is inclined $\theta=41^{\circ}$ above the horizontal (the block's initial velocity is zero). The block is dragged a distance of $30.5 m$ along the incline with a constant acceleration of $0.25\frac{m}{s^2}$. The coefficient of kinetic friction between the block and the ground is $0.31$.

A couple of points, which may be of interest.

1) A $41°$ slope is very steep and generally would be too steep to walk up. That would be into the realms of rock climbing. Certainly if a non-climber was put on a $41°$ slope they would not be happy.

For example, this "slab" is about $45°$. So, slightly steeper than the one in the problem.



2) $75kg$ is about the average person's body mass. Pulling that up a any slope would be very difficult. And pulling that up a $41°$ would be almost impossible.

3) The final speed up the slope, even with that seemingly modest acceleration, is about $4m/s$, which is about $14 km/h$. That's full running speed (equivalent to a 3-hour marathon).

So, the problem is totally unrealistic. Although, it can obviously be fixed by replacing the person with a mechanical device.

But, here's my point. If you are faced with a real-life problem, how important is it to check that the data is realistic? Do you just blindly plug in the numbers you have been given? Or, do you question the data you are given?

 

[neilparker62]

Having found the tension force, I can find the work done by the person who's pulling the block:
$$ W = 672.91N \cdot 30.5m $$
$$ W = 20510.2968J $$

Quick question: can I solve the above part with the Work-Energy theorem?

In principle , yes:

$$ F_{net}Δx = ΔKE⇒F_{net}=\frac{ΔKE}{Δx} $$

From $F_{net}$ you can calculate T but essentially that is what you have done anyway. Perhaps it is worth noting that "Work energy" arises from Newton II in the following way:

$$ F_{net} = ma ⇒ F_{net}Δx=maΔx = m \frac{(v_f-v_i)}{t} \times \frac{(v_f+v_i)t}{2} = ½m({v_f}^2-{v_i}^2) $$

 

[neilparker62]

Check the units.

More reasonably, after 30.5m the rope breaks but the object continues up a little before sliding down. Same answer, though.

I'm not sure about this: if the block continues sliding after the rope breaks, it will lose energy (due to friction) in the course of sliding upward and then back to the 30.5m mark. Also the OP should note that when the block slides down the slope, the parallel (to slope) component of weight and friction force will be acting in opposite directions.

 

[haruspex]

I'm not sure about this: if the block continues sliding after the rope breaks, it will lose energy (due to friction) in the course of sliding upward and then back to the 30.5m mark.

So? I'm saying it gives the same answer as interpretation 2 in post #2.

 

[neilparker62]

So? I'm saying it gives the same answer as interpretation 2 in post #2.

Apologies - had thought you meant same answer as for interpretation 1.