Question

Matt has a mass of 37 kg and skis down a hill with no friction or air resistance. The hill has an angle of 24 degrees. The free-body diagram is show below.

a. Write the expression for the net force in the x-direction.

b. Write the expression for the net force in the y-direction.

c. What is the acceleration in the y-direction?

d. What is the normal force acting on him?

Answer 1

a) mg cos(24)-N=0

b) mg sin(24)=ma

c) Since there is no original figure, in my coordinates a.y=0.

d) According to a): N=331.25 N

Answer 2

a) Along X direction since it can not move so Net force in x direction must be ZERO

`F_(net) = F_n - mgcos\theta`

since a = 0

`F_n - mgcos\theta = 0`

b)For Y direction net force is only due to component of the weight of object along the inclined plane

`mgsin\theta = F_(net)`

`m a_y = mgsin\theta`

c) for finding acceleration in Y direction we can use above equation

`m a_y = mg sin\theta`

divide both sides by mass

`a_y = g sin\theta`

`a_y = 9.8 sin24 = 3.98 m/s^2`

d) For finding Normal force we can use the expression of force in X direction

`0 = F_n - mg cos\theta`

`0 = F_n - (37* 9.8* cos24)`

`0 = F_n - 331.25`

`F_n = 331.25 N`