A 1.5-kg block slides at rest starts sliding down a snow-covered hill Point A, which has an altitude of 10 m. There is no friction on hill. After leaving the hill at point B, it travels horizontally toward a massless spring with force constant of 200 N/m. While travelling, it encounters a 20-m patch of rough surface CD where the coefficient of kinetic friction is 0.15. (a) What is the speed of the block when it reaches point B

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2020-07-19T05:30:59+02:00

Answer:

the speed of the block when it reaches point B is 14 m/s

Explanation:

Given that:

mass of the block slides = 1.5 - kg

height = 10 m

Force constant = 200 N/m

distance of rough surface patch = 20 m

coefficient of kinetic friction = 0.15

In order to determine the speed of the block when it reaches point B.

We consider the equation for the energy conservation in the system which can be represented by:

[tex]\dfrac{1}{2}mv^2=mgh[/tex]

[tex]\dfrac{1}{2}v^2=gh[/tex]

[tex]v^2=2 \times g \times h[/tex]

[tex]v^2=2 \times 9.8 \times 10[/tex]

[tex]v=\sqrt{2 \times 9.8 \times 10[/tex]

[tex]v=\sqrt{196[/tex]

v = 14 m/s

Thus; the speed of the block when it reaches point B is 14 m/s