As shown below, a 2.000 kg mass is sliding down an inclined plane that makes an
ID: 1996975 • Letter: A
Question
As shown below, a 2.000 kg mass is sliding down an inclined plane that makes an angle of 25.00 degree, at measured from the horizontal. The mass has an initial velocity of 3.000 m/s directed down the inclined plane, as shown. Initially there is no friction between the inclined plane and the mass. Using Newton's Laws, the forces acting on the mass were determined to be as shown below (you do not have to show how this was done, or verify my results - I assume you could have done this by yourself by now and just did the work to save you tome Exam time): Show how you calculate the work done by the Gravitational Force when the block slides 0.400 m down the incline. Show how you calculate the work done by the Normal Force when the block slides 0.400 m down the incline. What it the net (total) amount of work done on the mass when it slides 0.400 m down the incline? Using your results from above and the Work Energy Theorem for a Single Point Particle (or Object that can be treated as a Point Particle) - NOT NEWTON'S LAW'S !, AND NOT POTENTIAL ENERGY! - show how you calculate the final velocity of the mass after it has slid 0.400m down the incline. Suppose the situation was exactly the same as depicted above except that there was also an unknown constant Force of Kinetic Friction between the sliding mass and the inclined plane. Starting with the same initial velocity as before, and after moving down the incline the same distance as before, the final velocity of the block was measured and found to be 3.250 m/s, directed down the incline at shown: Using the Work Energy Theorem for a Single Point Particle (or Object that can be treated as a Point Panicle) - NOT NEWTON'S LAWS!, AND NOT POTENTIAL ENERGY! - show how you calculate the magnitude of the Force of Kinetic Friction. Using your result above, show how you calculate the Coefficient of Kinetic Friction between the mass and the inclined plane.Explanation / Answer
Work done = F.d (dot product) ]
= |F||d| cos(theta)
where theta = angle between force and displacement.
(A) F = m g = 19.6 N
theta = 90 - 25 = 65 deg
W1 = 19.6 x 0.400 x cos65 = 3.31 J
(B) For normal force, theta = 90 deg
and cos90 = 0
hence W = 0
(c) Net work done = 3.31 + 0 = 3.31 J
(D) Net work done = change in KE
3.31 = 2 v^2 / 2 - 2(3^2) / 2
v = 3.51 m/s
(E) Work done by friction + work done by gravity + work done by normal = change in KE
( - f x 0.400) + (3.31) + (0 ) = 2 (3.25^2 - 3^2) / 2
- 0.4f + 3.31 = 1.56
f = 4.37 N
(F) us = f / N
and N = m g cos25 = 17.76 N
us = 4.37 / 17.76 = 0.19 ......Ans
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