How far will the mug slide along the countertop if the coefficient of friction is 0.728 once released?

B is the answer to this question. The force is friction should be considered and related to the normal force. The coefficient of friction is 0.728. 0.728 will be multiplied to the mug of the root beer which is 1.49 kg and the normal force which is 9.8 m/s/s.

The equation will look like this: (0.728) (1.49 kg) (9.8 m/s/s). You will get 10.6 n as the answer. This will help you compute for the n force and the equation will be 11.9 N - 10.6 n = 1.27 n. Now you need to solve for the kinetic energy when you already have the n force. The formula you will use is: keb = w = f d cos(theta) or (1.27 n) (1.42 m) cos(0 deg). The answer will be 1.80 J.

Take note that the mug will start to slow down because of the friction so you would need to compute that using this formula: w = change in ke f d cos(theta) or -1.80 j (10.6 n) (d) cos(180 deg). The answer to this will be -180 J. This will be computed along with the earlier answers: -1.80 j - (10.6 n) (d) = -1.80 j d = 0.170 m.

0.170 m-the force of friction is related to the normal force (= mg) and the coefficient of friction (0.728). the ffrict is ffrict = mufnorm = (0.728) (1.49 kg) (9.8 m/s/s) = 10.6 n this means that the net force from a to b is 11.9 n - 10.6 n = 1.27 n. the net force from b to c is 10.6 n. from a to b, the work done equals the kinetic energy change. so the kinetic energy at position b is keb = w = f d cos(theta) = (1.27 n) (1.42 m) cos(0 deg) = 1.80 j from b to c, the mug will lose this same amount of kinetic energy as friction works upon it to bring it to a stop. so the work done from b to c is -1.80 j. w = change in ke f d cos(theta) = -1.80 j (10.6 n) (d) cos(180 deg) = -1.80 j - (10.6 n) (d) = -1.80 j d = 0.170 m