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

Question

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?

in progress 0
Thanh Thu 2 months 2021-07-27T17:37:48+00:00 1 Answers 2 views 0

Answers ( )

    0
    2021-07-27T17:39:15+00:00

    Answer:

    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          

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )