In the figure shown below, a 520 g box is placed in front of a spring, with spri
ID: 1449275 • Letter: I
Question
In the figure shown below, a 520 g box is placed in front of a spring, with spring constant 5.5 N/cm, that is compressed a distance Delta x from its natural length. When the box is let go, the spring will push the box and it will slide up the frictionless hill of height H = 1.5 in. At the top of the hill, the ground is not frictionless. The coefficient of kinetic friction in this area is 0.15 and the coefficient of static friction is 0.25. The box will travel a distance d = 7.0 cm on the rough surface at the top of the hill before coming to a stop. (a) What is the total work that friction does in bringing the box to a stop? (b) Before entering the area with friction what is the kinetic energy of the box? (c) What is the total mechanical (potential plus kinetic) energy of the box at the top of the hill? (d) What is the total mechanical energy at the bottom of the hill (after leaving the spring)? (e) How much potential energy is stored in the spring when it is fully compressed? (f) What distance, Deltax is the spring initially compressed?Explanation / Answer
Here, we will use the principle of conservation of energy to determine the required values. It needs to be understood that at the initial moment, the system will have energy only as the potential energy of the spring which in turn will change to kinetic energy as it will come off the spring. Then this kinetic energy will again turn into kinetic energy and potential energy at the top of the hill. This energy will then be exhausted by the action of the frictional force as it will start to move on the surface with friction.
We will make use of the above understanding to solve each of the given problems as follows:
a.) The box moves over a distance of 7cm on the surface. Also the friction acting on it would be N where N is the normal force and is the coefficient of kinetic friction.
That is the work done by the friction over the distance of 7 cm would be given as:
W = F.S = -0.52 x 9.81 x 0.15 x 7 x 10^-2 = -5.35626 x 10^-2 J is the work done by the friction.
b.) By principle of conservation of energy we can say that the magnitude of the kinetic energy of the block just before entering the area with friction would be same as the work done by the frictional force in bringing it to stand still.
Therefore kinetic energy of the block just before entering the area with friction is 5.35626 x 10^-2 J
c.) For calculating the potential energy of the block, we will take the bottom horizontal surface as the reference level.
Hence the total energy = PE + KE = mgH + 5.35626 x 10^-2 J = 0.52 x 9.81 x 1.5 + 0.053563 J = 7.705363 J
d.) At the bottom surface the net mechanical energy will be same as that at the top of the hill, just that all of it will be present in the form of kinetic energy and the potential energy would be zero for we have chosen the bottom surface as the reference level.
Hence the total mechanical energy = 7.705363 J
e.) Here again the net mechanical energy would be in the form of PE of the spring. Therefore the potential energy stored in the spring would be: 7.705363 J
f.) For a spring the potential energy stored is given as 0.5kx2
Hence we have: 0.5(550)x^2 = 7.705363
or, X = 2.80195 cm is the required compression of the spring.
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