Question

A 3.00-kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determine the force constant of the spring, if the box compresses the spring 5.20 cm before coming to rest. N/m (b) Determine the initial speed the box would need in order to compress the spring by 1.60 cm.

Answer 1

Answer:

a) k= 3594,7 N/m

b) v= 0.55 m/s

Explanation:

• a)
• As the surface is horizontal, the only change in energy will be the change in kinetic energy, as the box comes to an stop after compressing the spring.
• As we know that the surface is frictionless also, this change in kinetic energy must be equal to the change in the elastic potential energy of the spring.
• So we can write the following equality:

       $\Delta K = \Delta U$

       where $\Delta K = \frac{1}{2}*m*v^{2}$

       and $\Delta U = \frac{1}{2} * k* \Delta x^{2}$

• Simplifying and replacing by the values, we get:

        $3.00 kg* (1.8 m/s)^{2} = k* (0.052 m) ^{2}$        

• Solving for k:

$k = \frac{3.00kg*(1.8m/s)^{2} }{(0.052m)^{2}} = 3594.7 N/m$

• k = 3594.7 N/m

• b)
• For this part, we can just apply the same equality, replacing the value of k by the one we got, and solving for the initial speed v:

        $v = \sqrt{\frac{k*\Delta x^{2}}{m} } = \sqrt{\frac{3594.7N/m*(0.016m)^{2} }{3.00kg}} = 0.55 m/s$

• v = 0.55 m/s