A 10 kg farm wagon is sitting at the top of a hill that makes a 370 angle with the horizontal when the brake suddenly fails. The cart rolls down the hill for 50 m and then hits a haystack. It plows 2.0 m into the stack before coming to rest. Determine The force the haystack exerts on the wagon?

F = 1475.75 N      

Explanation:

Given:-

- The mass of the wagon m = 10 kg

- The slope angle θ = 37°

- The initial velocity of wagon at top of hill, vi = 0 m/s

- The amount of distance it plows into haystack, s2 = 2.0 m

-The wagon rolls down the slope for distance, s1 = 50 m

Find:-

Determine The force the haystack exerts on the wagon?

Solution:-

- First we must note that the wagon rolls down the slope with a constant acceleration due to gravity ( g ) component acting down the slope. The acceleration ( a ) of the wagon can be given as:

                       a = g*sin ( θ ).

- Since, the acceleration of the cart is constant we can apply third kinematic equation of motion with initial velocity at top of hill vi = 0 m/s and the velocity " v1 " right before it plows into the haystack at the bottom of hill after traveling a distance of s1 = 50 meters.

                       v1^2 = vi^2 + 2*a*s1  

                       v1^2 = 0 + 2*g*sin ( θ )*s1  

                       v1^2 = 2*9.81*sin ( 37 )*50    

                       v1 = √590.381

                       v1 = 24.30 m/s

- The constant force exerted by the haystack ( F ) as the wagon plows the haystack with a velocity "v1" by a distance of s2 and comes to, final velocity vf = 0, a stop.

Apply principle of work-done energy:

- Where, work is done on the wagon by haystack for W = F*s2.

                    W = Δ K.E        

                    W = 0.5*m* ( vf^2 - v1^2 )

                    F*s2 = 0.5*m*( v1 )^2

                    F = 0.5*m*( v1 )^2 / s2

                    F = 0.5*10*590.30 / 2    

                    F = 1475.75 N          

(75°f − 32) × 5/9 = 23.889°c

its c i just checked it