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maniacp08
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A 3.0kg slides along a horizontal surface with a speed of 7.0m/s.
After sliding a distance of 2.0m, the block makes a transition to a ramp inclined at an angle of 40 degrees to the horizontal. The coefficient of kinetic friction between the block and the surfaces is .30.
Find
a) the speed of the block when it reaches the ramp
b)the distance the block slides along the inclined surface before coming momentarilly at rest(neglect any energy dissipated along the transition curve)
For part A I used
External Work = change in potential energy + change in kinetic energy + change in thermal energy
External work = 0
change in potential energy = 0
so it becomes
0 = change in kinetic energy + change in thermal energy
change in kinetic energy = 1/2 m Vf^2 - 1/2 m Vi^2
change in thermal energy = Uk * FN * displacement = Uk * m * g * 2
I plug in the numbers and I solve for Vf?
For Part B:
I can use the same equation but this time I am solving for height.
External Work = change in potential energy + change in kinetic energy + change in thermal energy
External Work = 0
change in potential energy = -mgh
change in kinetic energy = -1/2 m Vi^2
change in thermal energy = Uk * FN * displacement = Uk * m * g * 2
Solve for the height it reaches, then use trig to find the distance/hypotenuse.
Are these approaches correct? Thanks for helping.
After sliding a distance of 2.0m, the block makes a transition to a ramp inclined at an angle of 40 degrees to the horizontal. The coefficient of kinetic friction between the block and the surfaces is .30.
Find
a) the speed of the block when it reaches the ramp
b)the distance the block slides along the inclined surface before coming momentarilly at rest(neglect any energy dissipated along the transition curve)
For part A I used
External Work = change in potential energy + change in kinetic energy + change in thermal energy
External work = 0
change in potential energy = 0
so it becomes
0 = change in kinetic energy + change in thermal energy
change in kinetic energy = 1/2 m Vf^2 - 1/2 m Vi^2
change in thermal energy = Uk * FN * displacement = Uk * m * g * 2
I plug in the numbers and I solve for Vf?
For Part B:
I can use the same equation but this time I am solving for height.
External Work = change in potential energy + change in kinetic energy + change in thermal energy
External Work = 0
change in potential energy = -mgh
change in kinetic energy = -1/2 m Vi^2
change in thermal energy = Uk * FN * displacement = Uk * m * g * 2
Solve for the height it reaches, then use trig to find the distance/hypotenuse.
Are these approaches correct? Thanks for helping.