In the Figure below, we see a 200 gram block resting on the horizontal portion o
ID: 1590352 • Letter: I
Question
In the Figure below, we see a 200 gram block resting on the horizontal portion of a “track”, whose left end is curved upwards. The horizontal portion of the track is frictionless, except for a 80.0 cm long “rough” portion, shown in the Figure, where the coefficient of kinetic friction is 0.400 . The curved portion of the track is frictionless. The block is held against a spring (which has a stiffness constant of 1,000 N/m), compressing the spring by 10.0 cm. Then the block is released from rest. (NOTE: The block is not attached to the spring; it is just held against the end of the spring.) 13. How much work is done on the block by gravity as the block slides up to its highest point on the curved portion of the track? a. zero. b. 5.00J. c. 4.37J. d. 0.588J. e. 5.00J. 14. Through what vertical distance does the block rise as it slides up to its highest point on the curved portion of the track? a. 2.23mm. b. 2.23cm. c. 2.55m. d. 2.55m. e. 2.23m. 15. Find the change in the gravitational potential energy of the system as the block slides up to its highest point on the curved portion of the track. a. zero. b. 5.00J. c. +4.37J. d.0.588J. e.+5.00J.
Explanation / Answer
Initial energy=spring energy= 0.5 k x2 =0.5*1000*0.1*0.1= 5J
Energy lost due to Friction, Efri= uMgs= 0.4*0.2*9.8*0.8 =0.63J
Gain in potential energy= 5- Efri= 5-0.63=4.37J,
13)Work done by gravity= -Gain in potential energy= -4.37 J..........c option
14)Vertical distance= Mgh/Mg= 4.37/(0.2*9.8) =2.23m........e option
15)Change in gravitational potential energy is 4.37 J, found above.....c option