Question #182048

. A block slides down a ramp that has a length of 11.4 m. The block travels at a speed of 5.43 m/s at the bottom of the ramp and increased in temperature from 22.1°C to 22.3°C. The block is made of copper and has a specific heat capacity of 385 J/kg°C. What is the angle of the ramp?


1
Expert's answer
2021-04-19T17:10:51-0400

mg+N+Ffr=ma,m\vec{g}+\vec{N}+\vec {F_{fr}}=m\vec a,

ma=mgsinαμmgcosα,ma=mgsin\alpha-\mu mg cos\alpha,

μ=gsinαagcosα,\mu=\frac{gsin\alpha-a}{gcos\alpha},

v=at, l=at22    a=v22l,v=at,~l=\frac{at^2}{2}\implies a=\frac{v^2}{2l},

Afr=Q,A_{fr}=Q,

μmgcosαl=cmΔt,\mu mg\cos \alpha l=cm\Delta t,

(gsinαv22l)l=cΔt,(gsin\alpha-\frac{v^2}{2l})l=c\Delta t,

sinα=v2+2cΔt2gl=0.82,sin\alpha=\frac{v^2+2c\Delta t}{2gl}=0.82,

α=55°.\alpha=55° .


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS