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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

User Qschulz
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Answer:

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

Step-by-step 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:


(1)/(2)mv^2=mgh


(1)/(2)v^2=gh


v^2=2 * g * h


v^2=2 * 9.8 * 10


v=\sqrt{2 * 9.8 * 10


v=\sqrt{196

v = 14 m/s

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

User Paresh Mayani
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