Question

Matt has a mass of 37 kg and skis down a hill with no friction or air resistance. The hill has a slop of 24°. A. What is the normal force acting on himB. What is his acceleration down hill?C. Now assume there is friction. If the coefficient of kinetic friction between him and the hill is .31, what is his acceleration down the hill?

Answer 1

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:

`mg\cos (\theta)-N=0`

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:

`\begin{gathered} 37\ast9.8\cos (24)-N=0 \\ \\ 37\ast9.8(0.9135)-N=0 \\ \\ 331.25-N=0 \end{gathered}`

Add N to both sides:

`\begin{gathered} 331.25-N+N=0+N \\ \\ 331.25=N \\ \\ N=331.25N \end{gathered}`

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:

`a=g\sin \theta`

Thus, we have:

`\begin{gathered} a=9.8\sin 24 \\ \\ a=9.8(0.4067) \\ \\ a=3.99m/s^2 \end{gathered}`

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:

`Fg=m\ast a_g`

Where:

Fg is the force due to gravity

m is the mass of Matt

Thus, we have:

`Fg=37\ast3.99=147.5N`

Also, let's find the force due to fricton using the formula:

`F_f=uN`

Where:

u is the coeficient of friction = 0.31

N is the normal force

We have:

`F_f=0.31\ast147.5=45.7N`

Thus, we have the formula:

`F_g-F_f=m\ast a`

Let's solve for a:

`\begin{gathered} 147.5-45.7=37\ast a \\ \\ 101.8=37\ast a \\ \\ a=(101.8)/(37) \\ \\ a=2.75m/s^2 \end{gathered}`

Therefore, the acceleration assuming there is friction is 2.75 m/s^2

ANSWER:

`\begin{gathered} A\text{. 331.25 N} \\ \\ B.3.99m/s^2 \\ \\ \text{ C. 2.75 m/s}^2 \end{gathered}`