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A 8.25kg block is sent up a ramp inclined at an angle 38.5 from the horizontal....

Question

A 8.25kg block is sent up a ramp inclined at an angle 38.5 from the horizontal....

A 8.25kg block is sent up a ramp inclined at an angle 38.5 from the horizontal. It is given an initial velocity v0=15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is ?k=0.477 and the coefficient of static friction is ?s= 0.845. HOw far up the ramp (in the direction along the ramp) does the block go before it comes to a stop? _______m

Answers

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else)

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please help

Image text transcribed for accessibility: A 6.25-kg block is sent up a ramp inclined at an angle theta = 38.5 degree from the horizontal. It is given an initial velocity v0 = 15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is mu k = 0.477 and the coefficient of static friction is mu s = 0.845. How far up the ramp (in the direction along the ramp) does the block go before it comes to a stop?

Answer

By conservation of energy

KE - Wf = PE

.5mv^2 - umgcos(angle)d = mgh (h = dsin(angle)

.5(15)^2 - (.477)(9.8)(cos 38.5)(d) = (9.8)(d)(sin 38.5)

112.5 - 3.658d = 6.1d

d = 11.53 m


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