Question

A 100g block is initially compressing a spring 5.0 cm. The spring launches the block 50cm horizontally along the ground with a friction force of 15N. The block then slides up a frictionless ramp at 30 degrees reaching a 1.5m maximum height off the ground. What was the force constant of the spring

Answer 1

Answer:

7200 N/m

Explanation:

Metric unit conversion

100g = 0.1 kg

5 cm = 0.05 m

50 cm = 0.5 m

As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground

The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

$W_f = F_fs = 15*0.5 = 7.5 J$

Let g = 10 m/s2. The change in potential energy can be calculated as the following:

$E_p = mgh = 0.1*10*1.5 = 1.5 J$

Therefore, as elastic energy is converted to potential energy and work of friction:

$E_e = W_f + E_p$

$kx^2/2 = 7.5 + 1.5 = 9 J$

$k = 9*2/x^2 = 18/0.05^2 = 7200 N/m$