Find the speed of the package in Figure at the end of the push, using work and e

Question

Find the speed of the package in Figure at the end of the push, using work and energy concepts. Here the work-energy theorem can be used, because we have just calculated the net work. W_net, and the initial kinetic energy, 1/2 mv_0^2. These calcinations allow us to find the final kinetic energy. 1/2 mv^2, and thus the final speed v. The work-energy theorem in equation form is W_net = 1/2 mv^2 - 1/2 mv_0^2. Solving for 1/2 mv^2 gives 1/2 mv^2 = W_net + 1/2 mv_0^2. Thus. 1/2 mv^2 = 92.0 J + 3.75 J = 95.75 J. Solving for the final speed as requested and entering known values gives v = squareroot 2(95.75 J)/m = squareroot 191.5 kg m^2/s^2/30.0 kg/2(95.75 J) Discussion Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the sum of the initial kinetic energy and the net work done on the package. This means that the work indeed adds to the energy of the package. How far does the package in Figure coast after the push, assuming friction remains constant? Use work and energy considerations. Strategy We know that once the person stops pushing, friction will bring the package to rest. In terms of energy, friction does negative work until it has removed all of the package's kinetic energy. The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement; hence, this gives us a way of finding the distance traveled after the person stops pushing.

Explanation / Answer

1/2mv^2 is final kinetic energy which is to be calculate from work done - initial kinetic energy and which is come to 95.75J,   3.75J is initial kinetic energy, and finally from the final kinetic energy velocity is being calculated.

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