A 0.5 kg mass moves up 40 centimeters along the incline shown in the figure belo
ID: 1361977 • Letter: A
Question
A 0.5 kg mass moves up 40 centimeters along the incline shown in the figure below. The vertical height of the incline is 7 centimeters.
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?
The mass is released from the top of the incline and slides down the incline. The maximum velocity (taken the instant before the mass reaches the bottom of the incline) is 1.06 m/s. What is the kinetic energy at that time?
What percentage of the initial potential energy has been converted into heat instead of kinetic energy?
Explanation / Answer
work done by the force is
W = F x d
= ( 1.0 N ) ( 40x10-2 m)
= 0.4 J
kinetic energy of the mass is
KE = ( 1/ 2) m v2
= ( 0.5 ) (0.5 kg ) (1.06 m/s)2
= 0.2809 J
potential energy of the mass is
PE = m gh
= ( 0.5 kg ) ( 9.8 m/s2) ( 7x10-2 m)
=0.343 J
energy released in the form of heat is
E = PE - KE
= 0.343 J - 0.2809 J
=0.0611 J
percentage of PE released in the form of heat is
= ( 0.0611 J / 0.343 J ) 100 = 18%
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.