Sunday, October 26, 2014

Problems

% latex2html id marker 815 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A box of mass 5.0 kg is pulled vertically upwards by a force of 68 N applied to a rope attached to the box. Find a) the acceleration of the box and b) the vertical velocity of the box after 2 seconds.
Solution:



 
Figure 4.1: Problem 4.1
\begin{figure} \begin{center} \leavevmode \epsfysize=4 cm \epsfbox{fig41.eps}\end{center}\end{figure}

a)
From the 2nd Law:
ma = T - mg   
$\displaystyle\Rightarrow$ a = $\displaystyle{\frac{T}{m}}$ - g   
  = $\displaystyle{\frac{68 \;{\:\rm N}}{5 \;{\:\rm kg}}}$ - 9.8 m/s 2 = 3.8 m/s 2  
b)
Since a is constant
v = v0 + at   
  = 0 + 3.8(2) = 7.6  m/s.  







% latex2html id marker 823 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A hockey puck of mass .5 kg travelling at 10 m/s slows to 2.0 m/s over a distance of 80 m. Find a) the frictional force acting on the puck and b) the coefficient of kinetic friction between the puck and the surface.
Solution:


 
Figure 4.2: Problem 4.2
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a)
First we find the acceleration of the puck from the kinematic equations of motion. We have, v0 = 10  m/s , v = 2  m/s and x = 80  m . The third equation of motion gives,
v 2 = v02 + 2ax   
$\displaystyle\rightarrow$ a = $\displaystyle{\frac{v^2-v_0^2}{2x}}$ = $\displaystyle{\frac{4-100}{2(80)}}$ = - 0.6  m/s 2  
From the Second Law:
In the x-direction,
fk = ma = .5(- 0.6) = - 0.3  N.       
b)
Use fk = - $\mu_{k}^{}$N . From the y-component of the 2nd Law: N - mg = 0 . Combining,
- $\displaystyle\mu_{k}^{}$mg = ma   
$\displaystyle\rightarrow$ $\displaystyle\mu_{k}^{}$ = - a/g   
  = 0.061.  







% latex2html id marker 840 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A student of mass 50 kg tests Newton's laws by standing on a bathroom scale in an elevator. Assume that the scale reads in newtons. Find the scale reading when the elevator is a) accelerating upward at .5 m/s 2, b) going up at a constant speed of 3.0 m/s and c) going up but decelerating at 1.0 m/s 2.
Solution:


 
Figure 4.3: Problem 4.3
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From the 2nd Law:

Fs - mg = ma   
$\displaystyle\rightarrow$ Fs = m(g + a).  
This gives:
a)
Fs = 50(9.8 + 0.5) = 515  N
b)
Fs = 50(9.8 + 0) = 490  N
c)
Fs = 50(9.8 - 1.0) = 440  N







% latex2html id marker 849 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A wooden plank is raised at one end to an angle of 30 o . A 2.0 kg box is placed on the incline 1.0 m from the lower end and given a slight tap to overcome static friction. The coefficient of kinetic friction between the box and the plank is $\mu_{k}^{}$ = 0.20 . Find a) the rate of acceleration of the box and b) the speed of the box at the bottom. Assume that the initial speed of the box is zero.
Solution:


 
Figure 4.4: Problem 4.4
\begin{figure} \begin{center} \leavevmode \epsfysize=3 cm \epsfbox{fig44.eps}\end{center}\end{figure}

a)
Find the components of the weight of the object:
wx = - mgsin $\displaystyle\theta$   
wy = - mgcos $\displaystyle\theta$.   
Write out the two components of Newton's 2nd Law:
  x : - mgsin $\displaystyle\theta$ + fk = - ma   
  y : N - mgcos $\displaystyle\theta$ = 0.  
Using fk = $\mu_{k}^{}$N we get,
ma = - $\displaystyle\mu_{k}^{}$(mgcos $\displaystyle\theta$) + mgsin $\displaystyle\theta$   
$\displaystyle\rightarrow$ a = g(sin $\displaystyle\theta$ - $\displaystyle\mu_{k}^{}$cos $\displaystyle\theta$)   
  = 9.8(sin 30 - 0.2cos 30) = 3.20 m/s 2  
b)
Since a is constant and v 2 = v02 + 2ax . With x = 1  m , v0 = 0 we have
v = $\displaystyle\sqrt{2(3.2)(1)}$ = 2.53  m/s .        







% latex2html id marker 876 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A 10 kg box is attached to a 7 kg box which rests on a 30 o incline. The coefficient of kinetic friction between each box and the surface is $\mu_{k}^{}$ = .1 . Find a) the rate of acceleration of the system and b) the tension in the rope.
Solution:


 
Figure 4.5: Problem 4.5
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We apply the 2nd law separately to each box.
For the 10 kg box:
y direction:

N2 - m2g =     
N2 = m2g,   
x direction:
T - fk 2 = m2a   
T - $\displaystyle\mu_{k}^{}$N2 = m2a   
$\displaystyle\rightarrow$ T - $\displaystyle\mu_{k}^{}$m2g = m2a. (13)
For the 7 kg box:
y direction:
N1 = m1gcos $\displaystyle\theta$,        
x direction:
m1gsin $\displaystyle\theta$ - T - fk 1 = m1a   
m1gsin $\displaystyle\theta$ - T - $\displaystyle\mu_{k}^{}$N1 = m1a   
$\displaystyle\rightarrow$ m1gsin $\displaystyle\theta$ - T - $\displaystyle\mu_{k}^{}$m1gcos $\displaystyle\theta$ = m1a.  
We have a system of two equations and two unknowns: a and T . We can solve as follows.

a)
From equation (4.1), T = m2a + $\mu_{k}^{}$m2g . Substituting into equation (4.2) gives,
m1a = m1gsin $\displaystyle\theta$ - $\displaystyle\mu_{k}^{}$m1gcos $\displaystyle\theta$ - m2a - $\displaystyle\mu_{k}^{}$m2g   
$\displaystyle\rightarrow$ m1a + m2a = m1gsin $\displaystyle\theta$ - $\displaystyle\mu_{k}^{}$m2g - $\displaystyle\mu_{k}^{}$m1gcos $\displaystyle\theta$  

$\displaystyle\rightarrow$ a = $\displaystyle{\textstyle\frac{1}{m_1+m_2}}$[m1gsin $\displaystyle\theta$ - $\displaystyle\mu_{k}^{}$m2g - $\displaystyle\mu_{k}^{}$m1gcos $\displaystyle\theta$]g$\displaystyle\mu_{k}^{}$]   
  = $\displaystyle{\textstyle\frac{1}{17}}$[7(9.8)sin 30 - (0.1)(10)(9.8) - (0.1)(7)(9.8)cos 30)] = 1.1 m/s 2  
b)
Then substituting into the first equation gives,
T = m2(a + g$\displaystyle\mu_{k}^{}$)   
  = 10(9.8(.1) + 1.1) = 20.8  N.
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