A farmer and his friend decide to move a heavy box from the loft of the farmer\'

Question

A farmer and his friend decide to move a heavy box from the loft of the farmer's barn. The farmer decides to tie a rope to a rafter in the barn and the other end to the box. The farmer then tells his friend to slide the box off of the loft. The farmer plans to catch it at the bottom of the swing. The box has a mass of 67.3 kg. The rope holding the box is 3.42 meters long and it barely reaches the box on the edge of the loft, making an angle of 56.0° with respect to vertical.

The box swings down to where the farmer is waiting. The farmer has a mass of 84.7 kg, and he is standing directly under the rafter where the box is tied. His center of mass coincides with the box's center of mass when it is hanging straight downward. He tries to catch the box, but instead he and the box swing upward.

a) How fast is the box swinging when it encounters the farmer?

b) How far from vertical does the box and the farmer swing after the box hits him?

c) How much mehcincal energy is lost in the process of the box hitting the farmer?

Explanation / Answer

A) let speed will be v. So by energy conservation

0.5*m*v*v = m*g*h => v = sqrt(2*g*h) = sqrt(2*9.8*3.42*(1-cos(56))) = 5.435 m/s Answer

B) velocity of both after collision :-

Applying momentum conservation

so 67.3*5.435 = (67.3+84.7)u

=> u = 67.3*5.435 / (67.3+84.7) = 2.406

Now applying energy conservation

0.5*m*u*u = m*g*h

=> h = 0.5*2.406*2.406/9.8 = 0.295

=> h = R(1-cos) => cos = 1 - h/R = 1 - 0.295/3.42 = 0.913

=> = acos(0.913) = 23.97 degree Answer

C) loss = niitial kinetic energy - final kinetic energy = 0.5*67.3*5.435*5.435 - 0.5*152*2.406*2.406 = 554.04 Joule Answer

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