Respuesta :
Given:
Mass of object = 37 kg
Slope = 24 degrees
Let's solve the following questions.
A. Normal force acting on him.
To find the normal force acting on Matt, apply the formula below:
[tex]mg\cos (\theta)-N=0[/tex]Where N is the force.
m is the mass = 37 kg
g is the gravitational acceleration = 9.8 m/s^2
Θ = 24 degrees.
Thus, we have:
[tex]\begin{gathered} 37\ast9.8\cos (24)-N=0 \\ \\ 37\ast9.8(0.9135)-N=0 \\ \\ 331.25-N=0 \end{gathered}[/tex]Add N to both sides:
[tex]\begin{gathered} 331.25-N+N=0+N \\ \\ 331.25=N \\ \\ N=331.25N \end{gathered}[/tex]Therefore, the normal force acting on Matt is 331.25 N
B. Acceleration down the hill.
The acceleration down the hill will be the opposite side(side opposite the angle).
To find the acceleration down the hill, apply the formula below:
[tex]a=g\sin \theta[/tex]Thus, we have:
[tex]\begin{gathered} a=9.8\sin 24 \\ \\ a=9.8(0.4067) \\ \\ a=3.99m/s^2 \end{gathered}[/tex]Therefore, the acceleration down the hill is 3.99 m/s
C. Given:
Coefficient of friction between Matt and the hill is = 0.31
Let's find the acceleration assuming there is friction.
To find the acceleration, we have the formula:
[tex]Fg=m\ast a_g[/tex]Where:
Fg is the force due to gravity
m is the mass of Matt
Thus, we have:
[tex]Fg=37\ast3.99=147.5N[/tex]Also, let's find the force due to fricton using the formula:
[tex]F_f=uN[/tex]Where:
u is the coeficient of friction = 0.31
N is the normal force
We have:
[tex]F_f=0.31\ast147.5=45.7N[/tex]Thus, we have the formula:
[tex]F_g-F_f=m\ast a[/tex]Let's solve for a:
[tex]\begin{gathered} 147.5-45.7=37\ast a \\ \\ 101.8=37\ast a \\ \\ a=\frac{101.8}{37} \\ \\ a=2.75m/s^2 \end{gathered}[/tex]Therefore, the acceleration assuming there is friction is 2.75 m/s^2
ANSWER:
[tex]\begin{gathered} A\text{. 331.25 N} \\ \\ B.3.99m/s^2 \\ \\ \text{ C. 2.75 m/s}^2 \end{gathered}[/tex]