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mojhsa [17]
3 years ago
14

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 b

y 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.
Physics
1 answer:
Vilka [71]3 years ago
7 0

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

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