a) [tex]0.4 m/s^2[/tex]
We can find the acceleration of the box by using the suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s is the distance travelled
u is the initial velocity
t is the time
a is the acceleration
For the box in the problem,
s = 16 m
t = 9 s
u = 0 (it starts from rest)
Solving for a, we find the acceleration:
[tex]a=\frac{2s}{t^2}=\frac{2(16)}{9^2}=0.4 m/s^2[/tex]
C) Tension force in the chain: 446 N
In order to find the normal force, we have to write the equation of the forces along the vertical direction.
We have 3 forces acting along this direction on the box:
- The normal force, N, upward
- The force of gravity, mg, downward (where m= mass of the box and g = acceleration of gravity)
- The vertical component of the tension of in the chain, [tex]T sin \theta[/tex], upward
So the equation of the forces along the vertical direction is
[tex]N+Tsin \theta - mg = 0[/tex] (1)
Along the horizontal direction, instead, we have the following equation:
[tex]T cos \theta - \mu N = ma[/tex] (2)
where
[tex]T cos \theta[/tex] is the horizontal component of the tension in the chain
[tex]\mu N[/tex] is the frictional force
a is the acceleration
From (1) we write
[tex]N=mg-T sin \theta[/tex]
And substituting into (2),
[tex]T cos \theta - \mu (mg-T sin \theta) = ma\\T cos \theta - \mu mg + \mu T sin \theta = ma\\T = \frac{ma+\mu mg}{cos \theta + \mu sin \theta}[/tex]
And substituting:
m = 100 kg
[tex]\theta=25^{\circ}[/tex]
[tex]\mu=0.46[/tex]
[tex]a=0.4 m/s^2[/tex]
[tex]g=9.8 m/s^2[/tex]
We find the tension in the chain:
[tex]T = \frac{(100)(0.4)+(0.46)(100)(9.8)}{cos 25 + 0.46 sin 25}=446 N[/tex]
B) Normal force: 792 N
We can now find the normal force by using again equation (1):
[tex]N+Tsin \theta - mg = 0[/tex]
And substituting:
T = 446 N
m = 100 kg
[tex]\theta=25^{\circ}[/tex]
[tex]g=9.8 m/s^2[/tex]
We find:
[tex]N=mg-T sin \theta=(100)(9.8)-(446)(sin 25)=792 N[/tex]
