A horizontal block-spring system with the block on a frictionless surface has to

Question

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 55.2 J and a maximum displacement from equilibrium of 0.198 m.

(a) What is the spring constant?
Correct: Your answer is correct. N/m

(b) What is the kinetic energy of the system at the equilibrium point?
Correct: Your answer is correct. J

(c) If the maximum speed of the block is 3.45 m/s, what is its mass?
Correct: Your answer is correct. kg

(d) What is the speed of the block when its displacement is 0.160 m?
Correct: Your answer is correct. m/s

(e) Find the kinetic energy of the block at x = 0.160 m.
Incorrect: Your answer is incorrect.
Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. J

(f) Find the potential energy stored in the spring when x = 0.160 m.
Correct: Your answer is correct. J

(g) Suppose the same system is released from rest at x = 0.198 m on a rough surface so that it loses 16.6 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant?

Explanation / Answer

The displacement is not given and the unit for energy is not given. Assuming that the maximum displacement is 0.214 m and total energy is 45 J (a) the spring constant k = 2E/ x² = 2*45/0.214² = 1965 N/m (b) the kinetic energy of the system at the equilibrium point is the total energy = 45 J (c) If the maximum speed of the block is 3.45 m/s, what its mass is m = 2E/ v² = 90/ 3.45² = 7.56 kg (d) The speed of the block when its displacement is 0.160 m is found from 0.5mv² = E – 0.5 k x² 0.5mv² = 45 – 0.5*1965*0.160 ² = 19.848 v² = 19.848 / (0.5*7.56)v = 2.29 m/s (e The kinetic energy of the block at x = 0.160 m is 0.5mv² = 19.9 J (f) The potential energy stored in the spring when x = 0.160 m is ( 45 - 19.9) = 25.1 J take E=45 (g) If the system is released from rest at x = 0.214 m on a rough surface so that it loses 14.4 J by the time it reaches its first turning point (after passing equilibrium at x = 0). its position at that instant is found as followed. Initial total energy = 45 J At the turning point its total energy is 45 – 14.4 = 30.6 J At the turning point its kinetic energy is zero. Hence its potential energy = 0.5k x² = 30.6 J x² = 30.6 / (0.5 *1965) x = 0.177 m

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