A box of 70kg is pulled up an incline measured 40 degrees from the horizontal po
ID: 2004938 • Letter: A
Question
A box of 70kg is pulled up an incline measured 40 degrees from the horizontal positive x axis, the rope parallel the x axis. The coefficient of friction is .4, and the box accelerates up the ramp at an acceleration of 1.5 m/s^2. The box starts at rest.
I calculated the tension of the rope to be 756.155N, the normal force to be 525.506N, and the friction force to be 210.202N. The weight force is (9.8 m/s^2)(70kg)
1. How much work does each force perform on the box while the box moves 5 meters?
2. What is the net work done on the box?
3. What is the final velocity of the box, using the equation of work energy theorem?
Explanation / Answer
(a) Work done due to tension, W1 = T * d = 756.155 * 5 = 3780.775 J Work done by the normal force, W2 = N * d * cos 90 = 0 J Work done by the frictional force, W3 = - f * d = - 210.20 * 5 = - 1051 J Work done by gravitational force, W4 = - W * h = - W * d sin 40 = - 70 * 9.8 * sin 40 = - 440.9 J (b) Net work done, Wnet = W1 + W2 + W3 + W4 = 3780.775 + 0 - 1051 - 440.9 = 2288.8 J (c) Work energy theorem: Change in kinetic energy = Net work done (1/2) m ( vf^2 - vi^2 ) = Wnet (1/2) * 70 * (vf^2 - 0) = 2288.8 vf = 8.1 m/sRelated Questions
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