A block of mass m begins at rest at the top of a ramp at elevation h with whatev

Question

A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height.

The block slides down the ramp over a distance d until it reaches the bottom of the ramp. How much of its original total energy (in J) survives as KE when it reaches the ground?

(In other words, the acceleration is not zero like it was in lab and friction does not remove 100% of the original PE. How much of that original energy is left over after the friction does work to remove some?)

m = 6.6 kg
h = 1.7 m
d = 5 m
= 0.3
= 36.87°

Answer:

If the leftover energy in the previous problem is 128.2 J (it's not, don't go back and try to use this value) and the mass is 2 kg, what speed (in m/s) does the block have at the bottom of its slide?

Answer:

A car moving at some speed hits the brakes and skids to a stop after 13 m on a level road. If the coefficient of friction for the road conditions of dry concrete is 0.53, what was the car's original speed (in m/s) before braking?

Answer:

Explanation / Answer

(a) Conservation of energy

Initial PE = m g h = 6.6 * 1.7 * 9.81 = 110.068 J

Energy Lost = m g cos(theta) d = 0.3 * 6.6 * 9.81 * 5 = 97.119 J

Amount of original energy left = 110.068 - 97.119 = 12.949 J

(b) E = 0.5 m v^2

v = sqrt( 2 E / m )

v= sqrt( 2 * 128.2 / 2 ) = 11.32 m/s

speed of tthe block have at the bottom of its slide =  11.32 m/s

(c) S = 13 m

= 0.53

v^2 = 2 a S

v^2 = 2 g S

v = sqrt( 2 g S )

v= sqrt( 2 * 0.53 * 9.81 * 13 ) = 11.626 m/s

car's original speed = 11.626 m/s

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