A 1.5 kg box that is sliding on a frictionless surface with a speed of 13 m/s ap
ID: 582287 • Letter: A
Question
A 1.5 kg box that is sliding on a frictionless surface with a speed of 13 m/s approaches a horizontal spring, (see the fioure (Figure 1)) The spring has a spring constant of 2100 N/m If one end of the spring is fixed and the other end changes its position, how far will the spring be compressed in stopping the box? Express your answer using two significant figures. How far will the spring be compressed when the box's speed is reduced to half of its initial speed? Express your answer using two significant figures.Explanation / Answer
A)
Apply conservation of energy
final potentail energy in the spring = initial kinetic energy of the block
(1/2)*k*x^2 = (1/2)*m*v^2
x^2 = m*v^2/k
==> x = v*sqrt(m/k)
= 13*sqrt(1.5/2100)
= 0.347 m
B)Apply conservation of energy
loss of kinetic energy = gain in potentail energy
0.5*m*(vi^2 - vf^2) = 0.5*k*x^2
x = sqrt(m*(vi^2 - vf^2)/k)
= sqrt(1.5*(13^2 - 6.5^2)/2100)
= 0.3 m
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