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A box 21.6 kg box initially at rest is pushed 2.36 along a rough horizontal floor, with a constant applied horizontal force of 56.2958 N. The acceleration of gravity is (9.8m/s^2). If the coefficient of friction between the box and the floor is 0.188,

1.find the work done by the friction. answer in J.

2. Find the final speed of the box in units of m/s

2007-10-14 11:15:21 · 1 answers · asked by contraseña 1 in Science & Mathematics Physics

1 answers

1. I assume you mean the box is pushed 2.35 m. Then the work done by friction is just umgx, where u is the coefficient of friction, m is the mass of the box, g is the acceleration due to gravity, and x is the distance the box was pushed; you were given all of these values.

2. The work you put into the box is Fx, where F is the applied horizontal force and x is the distance it was pushed; you were given both values. From this, subtract the work done by friction (which you just calculated), which opposed motion. The result is the net energy that was put into the box, which will be equal to Fx - umgx. This will be equal to the kinetic energy of the box at the end of observation. Kinetic energy is KE = mv^2, so v = sqrt(KE/m). And since we already had a result for what kinetic energy was equal to, we can say v = sqrt((Fx - umgx)/m), which you can calculate.

2007-10-17 10:07:08 · answer #1 · answered by DavidK93 7 · 0 0

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