a) 57.4 N
b) [tex]3.1 m/s^2[/tex]
c) 1.97 s
d) 13.7 N
e) 68.6 N
Explanation:
The text and some of the data in the problem are missing, so I will make some assumptions on the problem and on some values.
a)
Here there are two girls that are applying each a horizontal force of 30 N through two ropes on the sled, at an angle of [tex]25^{\circ}C[/tex] from each other.
Here we want to find the net tension force: therefore, we have to resolve the two forces along two perpendicular directions, and calculate the resultant.
Calling the two forces [tex]F_1,F_2[/tex] , and choosing the directions parallel and perpendicular to F1, we have:
- Along direction parallel to F1:
[tex]F_{1x}=30N\\F_{2x}=(30)(cos 30^{\circ})=26.0 N[/tex]
- Along the direction perpendicular to F1:
[tex]F_{1y}=0\\F_{2y}=(30)(sin 25^{\circ})=12.7 N[/tex]
Therefore, the components of the net force are
[tex]F_x=F_{1x}+F_{2x}=30+26.0=56.0 N\\F_y=F_{1y}+F_{2y}=0+12.7=12.7 N[/tex]
Therefore, the net tension force is:
[tex]T=\sqrt{F_x^2+F_y^2}=\sqrt{56.0^2+12.7^2}=57.4 N[/tex]
b)
The acceleration of the sled can be found by using Newton's second law of motion:
[tex]\sum F = ma[/tex]
where
[tex]\sum F[/tex] is the net force on the sled
m = 14 kg is the mass of the sled
a is the acceleration
The net force on the sled is given by the difference between the tension force T (forward) and the frictional force [tex]F_f=\mu mg[/tex] (backward), so we can write
[tex]T-\mu mg = ma[/tex]
where:
T = 57.4 N is the net tension
[tex]\mu=0.1[/tex] is the coefficient of friction
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
Solving for the acceleration,
[tex]a=\frac{T-\mu mg}{m}=\frac{57.4-(0.1)(14)(9.8)}{14}=3.1 m/s^2[/tex]
c)
The motion of the sled is a uniformly accelerated motion, so we can use the following suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
u is the initial velocity
t is the time
a is the acceleration
s is the displacement
In this problem, we have:
u = 0 (I assume the sled starts from rest)
s = 6.0 m (distance covered by the sled)
[tex]a=3.1 m/s^2[/tex] (acceleration)
Solving for t, we find the time needed:
[tex]t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(6.0)}{3.1}}=1.97 s[/tex]
d)
The force of kinetic friction on a moving object on a flat surface is given by
[tex]F_f = \mu mg[/tex]
where
[tex]\mu[/tex] is the coefficient of kinetic friction
m is the mass of the object
g is the acceleration due to gravity
For the sled in this problem, we have:
[tex]\mu=0.1[/tex]
m = 14 kg
[tex]g=9.8 m/s^2[/tex]
Therefore, the force of kinetic friction is:
[tex]F_f=(0.1)(14)(9.8)=13.7 N[/tex]
e)
Here the sled stops on a slope with angle of
[tex]\theta=30^{\circ}[/tex]
above the horizontal (I assumed this value since it is not given).
When the sled stops, there are two forces acting on it along the direction parallel to the slope:
- The component of the weight of the sled along the slope, down along the slope, of magnitude [tex]mg sin \theta[/tex]
- The force of static friction, [tex]F_f[/tex], up along the slope
The sled at this moment is in equilbrium, so the two forces are balanced; so we can write:
[tex]F_f = mg sin \theta[/tex]
And by substituting
[tex]m=14 kg[/tex]
[tex]g=9.8 m/s^2[/tex]
We find:
[tex]F_f=(14)(9.8)(sin 30^{\circ})=68.6 N[/tex]
Note: if some of the values that I have assumed are different, you can just plug in the correct numbers into the equation and find other results, but the procedure remains the same.
