How to find the speed of a block given a force against time graph?

Jun 2017
229
6
Lima, Peru
The problem is as follows:

A force $F$ whose modulus varies (shown in the figure from below) is applied to a block of $5\,kg$ of mass. This block slides over a flat rugous surface which has friction coefficients $\mu_s=0.3$ and $\mu_k=0.2$ respectively. It is known that the block begins moving from rest. Find approximately the speed in $\frac{m}{s}$ of the block when $t=5\,s$. You may use the value of the gravity $g=9.81\,\frac{m}{s^2}$.



The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&1.54\,\frac{m}{s}\\
2.&2.19\,\frac{m}{s}\\
3.&3.22\,\frac{m}{s}\\
4.&4.16\,\frac{m}{s}\\
5.&5.24\,\frac{m}{s}\\
\end{array}$

I'm here lost at exactly how should I relate the graph given to get the necessary information in order to get the speed. The source of my confusion is that how exactly should I understand the word approximately speed?. Could it be that it is referred this way because is related with the average net force?.

Can somebody help me here please?.
 

skeeter

Math Team
Jul 2011
3,135
1,698
Texas
$\displaystyle \Delta p = m\Delta v = \int_{t_0}^{t_f} F_{net} \, dt$

$F_{net} = F(t) - f_k$

$\displaystyle m(v_f-v_0) = \int_0^5 F(t) - \mu_k mg \, dt = 60 - 9.81(5) = 10.95 \text{ kg m/s} \implies v_f = 2.19 \text{ m/s}$
 

topsquark

Math Team
May 2013
2,373
990
The Astral plane
The problem is as follows:

A force $F$ whose modulus varies (shown in the figure from below) is applied to a block of $5\,kg$ of mass. This block slides over a flat rugous surface which has friction coefficients $\mu_s=0.3$ and $\mu_k=0.2$ respectively. It is known that the block begins moving from rest. Find approximately the speed in $\frac{m}{s}$ of the block when $t=5\,s$. You may use the value of the gravity $g=9.81\,\frac{m}{s^2}$.



The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&1.54\,\frac{m}{s}\\
2.&2.19\,\frac{m}{s}\\
3.&3.22\,\frac{m}{s}\\
4.&4.16\,\frac{m}{s}\\
5.&5.24\,\frac{m}{s}\\
\end{array}$

I'm here lost at exactly how should I relate the graph given to get the necessary information in order to get the speed. The source of my confusion is that how exactly should I understand the word approximately speed?. Could it be that it is referred this way because is related with the average net force?.

Can somebody help me here please?.
What skeeter said.

Also, please note that the impulse given to an object is the area under the t vs. F graph. (The integral format that skeeter is using is the exact same principle.) In this case the given graph has been blocked out to show you how to do the area: two triangles and two rectangles.

-Dan
 
  • Like
Reactions: skeeter
Jun 2017
229
6
Lima, Peru
What skeeter said.

Also, please note that the impulse given to an object is the area under the t vs. F graph. (The integral format that skeeter is using is the exact same principle.) In this case the given graph has been blocked out to show you how to do the area: two triangles and two rectangles.

-Dan
If I were to follow that route the area to be considered is between $0$ to $5s$ which will be:

$A=\frac{5\left(16+8\right)}{2}=60\,Ns$

which correspond to what skeeter found for only one term in the integral. But what about the effect of the friction, should this part be calculated from an area?.
 
Jun 2017
229
6
Lima, Peru
$\displaystyle \Delta p = m\Delta v = \int_{t_0}^{t_f} F_{net} \, dt$

$F_{net} = F(t) - f_k$

$\displaystyle m(v_f-v_0) = \int_0^5 F(t) - \mu_k mg \, dt = 60 - 9.81(5) = 10.95 \text{ kg m/s} \implies v_f = 2.19 \text{ m/s}$
I'm still stuck at how did you obtained $60$?.

Does it imply you are integrating this function? (sorry maybe it not too obvious for me)

$F(t)=-\frac{8}{5}t+16$

Therefore integrating in the interval would give:

$\left[\frac{-4t^2}{5}+16t\right]^5_0=-20+80=60$

It was also mentioned that I could use finding the area by geometrical relationships, but what's exactly the justification for the area in the frictional force, is it seen in the graph?.

I'm referring to

$\left[\mu_k mg t\right]^5_0= 0.2 \times 5 \times 9.81 \times 5=49.05$

what would be its equivalent in the graph?.

Because I'm getting the idea that I'm subtracting two areas. Can you do a calculus-free version of the answer to compare both please?. :)
 
Last edited:

skeeter

Math Team
Jul 2011
3,135
1,698
Texas
Note the graph does not take friction force into account.

Since the applied force is linear ...

impulse due to the applied force = $F_{avg} \Delta t = \dfrac{16 + 8}{2} \cdot 5 = 60 \text{ kg m/s}$

and friction force is a constant ...

impulse due to friction = $-f_k \cdot \Delta t = -\mu_k \cdot mg \cdot \Delta t = -(0.2)(5 \cdot 9.81)(5) = -49.05 \text{ kg m/s}$

$\Delta p = m\Delta v = 60 - 49.05 = 10.95 \implies \Delta v = \dfrac{10.95 \text{ kg m/s}}{5 \text{ kg}}$
 
  • Like
Reactions: topsquark
Jun 2019
455
237
USA
In case it hasn't clicked yet, taking the integral of a function between two limits is identically the same as finding the area under the curve between those limits.

You can simplify a lot of problems by understanding this.
 
  • Like
Reactions: topsquark