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Homework answers / question archive / The height above the ground at position 10m is 1
The height above the ground at position 10m is 1.73 The simulation graph energy values (object is on the top of the ramp) Kinetic energy (top) 0.00 J Potential energy (top): 1701.75 J Total mechanical energy (top): 1701.75 The simulation graph energy values (object is on the bottom of the ramp) Kinetic energy (bottom): 1701.75 J Potential energy(bottom): 0.00 J Total mechanical energy(bottom): 1701.75 J
11. Calculate the speed of the filing cabinet at the bottom of the ramp, using equation 3 from the theory. 12. Use the distance the cabinet traveled down the ramp, its initial speed down the ramp, and the time it took the cabinet to go down the ramp (found in the top left corner) to also calculate final speed (use equations 5 and 6 from theory). 13. Compare the final speeds you got from #11 and #12. Explain the difference.
The answers are given below in the explanation.
Step-by-step explanation
(11) Given :
the kinetic energy of the cabinet at the bottom of the ramp = KEbottom = 1701.75 J,
the mass of the cabinet = m = 100 kg.
Hence, the speed of the cabinet at the bottom of the ramp is :
vf1 = √ ( 2 x KEbottom / m )
= √ ( 2 x 1701.75 J / 100 kg )
= 5.83 m / s.
(12) Given :
the distance the cabinet travelled down the ramp = d = 10 m,
its initial speed down the ramp = u = 0 m / s,
the time it took the cabinet to go down the ramp = t = 2.1 s.
Therefore, the acceleration of the cabinet down the ramp is :
a = 2d / t2
= 2 x 10 m / ( 2.1 s )2
= 4.535 m / s2.
Hence, the speed of the cabinet at the bottom of the ramp is :
vf2 = u + at
= 0 m / s + 4.535 m / s2 x 2.1 s
= 9.5 m / s.
(13) vf2 / vf1 = 9.5 / 5.83 ~ 1.6.
Hence, the final speed from #12 is almost 1.6 times the final speed from #11.
This difference comes because as the friction force is suddenly removed by clicking the " frictionless " button, and hence the cabinet accelerates at a faster rate to reach its final speed of 5.83 m / s.
