A 7.20-kg block of ice, released fromrest at the top of a 1.70-m-longfrictionles

Question

A 7.20-kg block of ice, released fromrest at the top of a 1.70-m-longfrictionless ramp, slides downhill, reaching a speed 2.45 m/s of at the bottom.
(a) What is the angle between the ramp and thehorizontal?
°

(b) What would be the speed of the ice at the bottom if the motionwere opposed by a constant friction force of 10.0 N parallel to thesurface of the ramp?
m/s (a) What is the angle between the ramp and thehorizontal?
°

(b) What would be the speed of the ice at the bottom if the motionwere opposed by a constant friction force of 10.0 N parallel to thesurface of the ramp?
m/s

Explanation / Answer

let the height from which the block was released=h m=mass of the block g=9.8 m/2 change in kineticenergy=(1/2)*m*2.452-0=(1/2)*m*2.452=potentialenergy (1/2)*m*2.452=m*g*h h=2.452/(2*9.8) h=0.31 m angle of the ramp=sin-1(0.31/1.7)=10.5 degree let when frictional force was acting speed at the bottom=v work done by friction=10*1.7=17 J initial potential energy=m*g*h=7.2*9.8*0.31=21.87 J change in potential energy=21.87-0=21.87 J net change in kinetic energy=21.87-17=4.87 J (1/2)*m*v2-0=4.87 (1/2)*7.2*v2=4.87 v=1.16 m/s

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