Respuesta :
Answer:
ΔEP = -1.36 J
Explanation:
Given:
- The complete question is as follows:
" A mass of 0.105 kg hangs from a vertical spring in the lab room. You pull down on the mass and throw it vertically downward. The speed of the mass just after leaving your hand is 5.20 m/s. "
Find:
When the mass has moved downward a distance of 0.04 m, the speed of the mass has decreased to 1.39 m/s.
Solution:
- The initial velocity of the system vi = 5.20 m/s
- The mass m = 0.105 kg
- The final velocity of mass vf = 1.39 m/s
- Change in distance d = 0.04m
Solution:
- When the mass (m) is moved down by (d) the work-done by gravity P.E translates to change in kinetic energy K.E and elastic potential energy of the spring EP . The energy balance can be set as:
ΔEP + P.E = ΔK.E
ΔEP + m*g*h = 0.5*m*(vf^2-vi^2)
ΔEP = 0.5*m*(vf^2-vi^2) - m*g*h
ΔEP = 0.105* (- 9.81*0.04 + 0.5*(1.39^2-5.2^2) )
ΔEP = - 1.36 J
- Since, The work done by spring is negative because the displacement is downwards and the force is upwards.
