Multiple choice, no details required or graded. A box of mass 2 kg is released f
ID: 1431329 • Letter: M
Question
Multiple choice, no details required or graded. A box of mass 2 kg is released from rest at the top of the ramp shown in the figure. The friction coefficient of the ramp is 0.1, the friction coefficient of the level surface is 0.05, and we use g = 10 m/s^2. The numbered circles indicate relevant situations for the problem: (1) when the box is at the top of the ramp, (2) when the box is at the bottom of the ramp (we assume a smooth transition from the ramp to the horizontal floor), and (3) at a distance s from the end of the ramp. Which of the following principle(s) apply when the box slides down the ramp? W_net =k_2-k_1 u_1+k_1 = u_2+k_2 u_1+k_1 = u_2+k_2-w_fr only(1) only(2) only (3) (1)and (2) (1) and (3) Calculate the work done by the friction force as the box slides down the ramp. The work due to friction is -23 J -20 J -12 J 20 J 120 J The speed of the box in situation (2), at the bottom of the ramp, is 10 m/s. What is the kinetic energy of the box in state (2)? 10 m/s 10 J 100 m^2/s^2 100 120 J Calculate the speed of the box in situation (3), after the box has traveled a distance of 19 m on a level surface with friction coefficient 0.05 0 m/s 2 m/s 4.5 m/s 9 m/s 10 m/sExplanation / Answer
a) only iii
b) Work by friction = - mg mu cos theta * distance - mg mu * distance = - 2 * 10 * 0.1 * (10 / sqrt(10^2 + 6^2)) - 2*10*0.05*19 = - 20.71 J
So the answer is -20 J (closest value)
c) Kinetic energy at 2 = 0.5*m*v^2 = 0.5*2*10^2 = 100 J
d) Loss of energy between 2 and 3 = mg * mu * distance = 2*10*0.05*19 = 19 J
So kinetic energy at 3 = 100 - 19 = 81 m
So the velocity v will satisfy; 81 = 0.5 * 2 * v^2
=> v = 9 m/s
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