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A 8.00 kg block of ice, released from rest at the top of a 1.50-m-long frictionless ramp, slides downhill, reaching a speed of 2.70 m/s at the bottom.What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.0 N parallel to the surface of the ramp

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okpalawalter8

We have that the speed of the ice at the bottom  is mathematically given as

v=1.8812m/s

The speed of the ice at the bottom

Question Parameters:

A 8.00 kg block of ice, released from rest

at the top of a 1.50-m-long frictionless ramp,

slides downhill, reaching a speed of 2.70 m/s

constant friction force of 10.0 N parallel to the surface of the ramp

Generally the equation for the Potential Energy is  is mathematically given as

P.E=K.E+W_{fric}

Therefore

mgh=0.5mv^2+W_{fric}

Where

W_{fric}=F.S

W_{fric}=10*1.50

W_{fric}=15

And

mgh=0.5mv^2+

h=0.3719

Hence

0.5(8.00)(v^2)=(8.00)(9.8)(0.3719)-15

v^2=3.53924

v=1.8812m/s

For more information on Velocity visit

https://brainly.com/question/862972

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