An ideal spring with a spring constant of 3000 N/m is attached at the bottom of a rough inclined plane that makes an angle of 37 degrees with the horizontal.The spring is compressed a distance of 80 cm and then a 12-kg box is placed on the spring.The coefficient of kinetic friction between the box and the inclined plane is 0.2.If the box is released from rest and is not attached to the spring,how fast will the box be traveling after sliding 4 m up the incline (3.2 m beyond the uncompressed length of the spring)?

Question

Quizplus · Accepted Answer

To solve this problem, we use the principle of conservation of energy, taking into account the work done against friction. The initial energy is the potential energy stored in the spring, given by $ \frac{1}{2}kx^2 $, where $ k = 3000 \, \text{N/m} $ is the spring constant and $ x = 0.8 \, \text{m} $ is the compression distance. The final energy is the kinetic energy of the box, $ \frac{1}{2}mv^2 $, where $ m = 12 \, \text{kg} $ is the mass of the box and $ v $ is its velocity. The work done against friction is $ f_k d = \mu_k mg \cos(\theta) d $, where $ \mu_k = 0.2 $ is the coefficient of kinetic friction, $ g = 9.8 \, \text{m/s}^2 $ is the acceleration due to gravity, $ \theta = 37^\circ $ is the angle of the incline, and $ d = 4 \, \text{m} $ is the distance the box slides up the incline.The energy equation becomes:$$ \frac{1}{2}kx^2 = \frac{1}{2}mv^2 + \mu_k mg \cos(\theta) d $$Plugging in the given values and solving for $ v $ yields a velocity of approximately $ 6.9 \, \text{m/s} $, making choice A correct.

An Ideal Spring with a Spring Constant of 3000 N/m