Question

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed by 11 cm from its uncompressed length. The compressed spring and block rests at the bottom of an incline of 28◦ with the spring lying along the surface of the ramp.
After all the external forces are removed (so the compressed spring releases the mass) how far D along the plane will the block move before coming to a stop? Answer in units of m.

Answer 1

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=$\frac{1}{2} kx^{2} +PE1$

Energy at the point where the block will stop consists of only gravitational potential energy=$PE2$

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒$\frac{1}{2} kx^{2} +PE1=PE2$

⇒$PE2-PE1=\frac{1}{2} kx^{2}$

Also $PE2-PE2=mgh$

where $m$ is the mass of block

$g$ is acceleration due to gravity=9.8 m/s

$h$ is the difference in height between two positions

⇒$mgh=\frac{1}{2} kx^{2}$

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴$2100*9.8*h=\frac{1}{2}*2200*0.11^{2}$

⇒$20580*h=13.31$

⇒$h=\frac{13.31}{20580}$

⇒h=0.0006467m=$6.5e-4$