How Far Does a Block Slide Up a Ramp Before Stopping?

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A block of mass 2.96 kg slides up a ramp at an initial speed of 12.8 m/s, inclined at 36.4 degrees, with a coefficient of kinetic friction of 0.45. The work-energy principle is applied, where the work done against friction and gravitational potential energy must be considered to determine the block's displacement before it stops. The gravitational potential energy depends on the height, which can be calculated using the ramp's angle. The discussion emphasizes the need to include gravitational work in the calculations for accurate results. Understanding the relationship between height and displacement along the ramp is crucial for solving the problem.
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Homework Statement


A block of mass 2.96 kg is sliding up a ramp with an initial speed of 12.8 m/s. The ramp is inclined from the horizontal at an angle of 36.4 degrees and the coefficient of kinetic friction is \mu=045. What is the displacement of the block along the ramp from the initial time until it stops?


Homework Equations



W=ΔE
W=∫F*ds
ΔE=(m/2)(vf)2-(m/2)(v0)2 where (vf)=0
\Phi=36.4
m=2.96kg
v0=12.8 m/s
\mu= 0.45

The Attempt at a Solution



\mu*m*g*cos\Phix=-(m/2)(v0)2
solve for x
x=-(m/2)(v0)2/\mu*m*g*cos\Phi
 
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hi getty102! :smile:

correct method, but you forgot to include the gravitational PE (or gravitational work done) :wink:
 
Cool, thank you. The potential energy from gravity is dependent on the height. Which in this problem is kind of what I'm looking for because if I can find the height I would know the total displacement of the block up the ramp.
 
xsinΦ ? :wink:
 
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