A force of 6.0 N pulls a box 0.40 m along a frictionless plane that is inclined at 36°. What work is being done by the pulling f
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<span>the spring constant is 200kN/m.
</span>
The work done on the spring by the the free falling safe is equal to the kinetic energy of the spring:
W1 =
k
The work done on the safe by gravity is the potential energy of the spring:
W2 = (mass of safe) x g x (height)
= 1000kg x 9.8 m/s^2 x (2 + 0.54)m
=1000kg x 9.8 m/s^2 x (2.54)m
= 24,892 N-m
W2 = W1 =24892 N-m
⇒k
= 24892 N-m
Therefore, k =
= 199136 kN/m.
The spring constant is 199,136 N/m ≈ 200 kN/m
</span>
The work done on the spring by the the free falling safe is equal to the kinetic energy of the spring:
W1 =
The work done on the safe by gravity is the potential energy of the spring:
W2 = (mass of safe) x g x (height)
= 1000kg x 9.8 m/s^2 x (2 + 0.54)m
=1000kg x 9.8 m/s^2 x (2.54)m
= 24,892 N-m
W2 = W1 =24892 N-m
⇒
Therefore, k =
The spring constant is 199,136 N/m ≈ 200 kN/m