A 10 kg block on a horizontal surface is attached to a horizontal spring of spri
ID: 2029398 • Letter: A
Question
A 10 kg block on a horizontal surface is attached to a horizontal spring of spring constant k = 5.6 kN/m. The block is pulled to the right so that the spring is stretched 8.9 cm beyond its relaxed length, and the block is then released from rest. The frictional force between the sliding block and the surface has a magnitude of 72 N. (a) What is the kinetic energy of the block when it has moved 2.1 cm from its point of release? (b) What is the kinetic energy of the block when it first slides back through the point at which the spring is relaxed? (c) What is the maximum kinetic energy attained by the block as it slides from its point of release to the point at which the spring is relaxed?I got the answer for part A is 19.432J, but it turned out wrong answer.
So I tried 15.504 J but still wrong answers.
Explanation / Answer
(1) Given that, mass of the block, m = 10 kg force constant, k = 5.6 kN/m = 5.6 x 103 N/m stretched length, x = 8.9 cm = 8.9 x 10-2 m restoring force of the spring, f = kx frictional force, fk = 72 N distance travelled, s = 2.1 cm = 2.1 x 10-2 m initial speed, u = 0 m/s The net force acting on the mass is, F = f - fk ma = kx - 72 (10 kg)a = (5.6 x 103 N/m)(8.9 x 10-2 m) - 72 a = 42.64 m/s2 From the kinematic relation, v2 - u2 = 2as v2 = 2 (42.64 m/s2) (2.1 x 10-2 m) = 1.79088 (m/s)2 Therefore, the kinetic energy of the block is, KE = (1/2)mv2 = (0.5) (10 kg)(1.79088 (m/s)2 KE = 8.9544 Jma = kx - 72 (10 kg)a = (5.6 x 103 N/m)(8.9 x 10-2 m) - 72 a = 42.64 m/s2 From the kinematic relation, v2 - u2 = 2as v2 = 2 (42.64 m/s2) (2.1 x 10-2 m) = 1.79088 (m/s)2 Therefore, the kinetic energy of the block is, KE = (1/2)mv2 = (0.5) (10 kg)(1.79088 (m/s)2 Therefore, the kinetic energy of the block is, KE = (1/2)mv2 = (0.5) (10 kg)(1.79088 (m/s)2 KE = 8.9544 J
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