A 0.6 kg mass lies on a horizontal surface. A spring whose spring constant 40 N/
ID: 1458250 • Letter: A
Question
A 0.6 kg mass lies on a horizontal surface. A spring whose spring constant 40 N/m is placed on the surface and compressed by 75 cm. The mass is placed against the spring and held at rest. The spring is then released and as the mass accelerates under the influence of the expanding spring, it experiences a constant frictional force of 10 N. a. What will be the kinetic energy of the mass when the spring is still compressed by 25cm.? b. What are the magnitudes of the net force and acceleration experienced by the mass when at the position in part (a)? c. What will be the total distance the mass travels, from its initial position (not where it loses contact with the spring) until it comes to a stop?Explanation / Answer
given,
mass = 0.6 kg
k = 40 N/m
x = 0.75 m
frictional force = 10 N
by conservation of energy
initial energy = final energy
0.5 * kx^2 = 0.5 * k * x1^2 + KE + frictional force * distance
0.5 * 40 * 0.75^2 = 0.5 * 40 * 0.25^2 + KE + 10 * (0.75 - 0.25)
KE = 5 J
kinetic energy of the mass = 5 J
net force = kx - frictional force
net force = 40 * 0.25 - 10
net force = 0 N
since the force is 0 the acceleration will be 0 m/s^2
speed of the block when it loses contact with spring v
0.5 * 40 * 0.75^2 = 0.5 * 0.6 * v^2 + 10 * 0.75
v = 3.535 m/s
acceleration = 10 / 0.6
v^2 = u^2 + 2as
0 = 3.535^2 - 2 * (10 / 0.6) * s
s = 0.3748 m
total distance covered = 0.3748 + 0.75
total distance covered = 1.1248 m
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