Question

100 POINTS, NEED HELP ASAP i just don’t get how to calculate all of these things

Answer 1

Answer:

a) 57.4 N

b) 31 m/s²

c) 1.97 s

d) 13.7 N

e) 68.6 N

Explanation:

Hope that helped!

Answer 2

a) 57.4 N

b) $3.1 m/s^2$

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 $25^{\circ}C$ 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 $F_1,F_2$ , and choosing the directions parallel and perpendicular to F1, we have:

- Along direction parallel to F1:

$F_{1x}=30N\\F_{2x}=(30)(cos 30^{\circ})=26.0 N$

- Along the direction perpendicular to F1:

$F_{1y}=0\\F_{2y}=(30)(sin 25^{\circ})=12.7 N$

Therefore, the components of the net force are

$F_x=F_{1x}+F_{2x}=30+26.0=56.0 N\\F_y=F_{1y}+F_{2y}=0+12.7=12.7 N$

Therefore, the net tension force is:

$T=\sqrt{F_x^2+F_y^2}=\sqrt{56.0^2+12.7^2}=57.4 N$

b)

The acceleration of the sled can be found by using Newton's second law of motion:

$\sum F = ma$

where

$\sum F$ 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 $F_f=\mu mg$ (backward), so we can write

$T-\mu mg = ma$

where:

T = 57.4 N is the net tension

$\mu=0.1$ is the coefficient of friction

$g=9.8 m/s^2$ is the acceleration due to gravity

Solving for the acceleration,

$a=\frac{T-\mu mg}{m}=\frac{57.4-(0.1)(14)(9.8)}{14}=3.1 m/s^2$

c)

The motion of the sled is a uniformly accelerated motion, so we can use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

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)

$a=3.1 m/s^2$ (acceleration)

Solving for t, we find the time needed:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(6.0)}{3.1}}=1.97 s$

d)

The force of kinetic friction on a moving object on a flat surface is given by

$F_f = \mu mg$

where

$\mu$ 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:

$\mu=0.1$

m = 14 kg

$g=9.8 m/s^2$

Therefore, the force of kinetic friction is:

$F_f=(0.1)(14)(9.8)=13.7 N$

e)

Here the sled stops on a slope with angle of

$\theta=30^{\circ}$

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 $mg sin \theta$

- The force of static friction, $F_f$, up along the slope

The sled at this moment is in equilbrium, so the two forces are balanced; so we can write:

$F_f = mg sin \theta$

And by substituting

$m=14 kg$

$g=9.8 m/s^2$

We find:

$F_f=(14)(9.8)(sin 30^{\circ})=68.6 N$

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.