A 2.0 kg rock is pressed up a distance delta x 20 cm against a spring with sprin
ID: 1419956 • Letter: A
Question
A 2.0 kg rock is pressed up a distance delta x 20 cm against a spring with spring constant 400 N/m and then released causing it to slide across a friction less bore mill surface and then to slide down a frictionless incline At the bot loin of the incline is a horizontal surface that has a coefficient of kinetic friction mu_i = 0.80 with the rock, which causes tin- rock to come to test in L 1.5 m. Fill out the Energy Chart below to calculate the Potential and Kinetic Energy at each time point. Use g 10 m s^2 for easier calcination Assume Energy Conserved UNTIL the rock reach the rough surface.Explanation / Answer
assume that initial height is h.
then initial gravitational potential energy =m*g*h=2*10*h=20*h Joules
initial spring potential energy=0.5*spring constant*compression^2=0.5*400*0.2^2=8 J
initial kinetic eenrgy =0 (as the rock is stationary)
hence initial total energy=20*h+8 joules
when the rock comes to the bottom of the incline, total potential energy=0
let speed be v m/s
then total kinetic energy=0.5*2*v^2=v^2 joules
as energy is conserved,
v^2=20*h+8....(1)
deceleration due to friction=friction coefficient*g=0.8*10=8 m/s^2
while travelling from B to C:
initial speed=v m/s
acceleration=-8 m/s^2
distance covered=1.5 m
final speed=0
using the formula:
final speed^2-initial speed^2=2*acceleration*distance
==>0^2-v^2=2*(-8)*1.5
==>v^2=24...(2)
using equation 2 in equation 1,
20*h+8=24
==>20*h=16
==>h=16/20=0.8 m
hence the table can be filled as below:
PE_gravity :
in scene A, h=0.8 m
==>PE_gravity=2*10*0.8=16 J
in scene B and scene C , height=0
==>PE_gravity=0
KInetic energy:
in scene A: speed=0==>kinetic energy=0
in scene B: kinetic energy=0.5*m*v^2=0.5*2*24=24 J
in scene C: speed=0==>KE=0
PE_spring:
in scene A, spring energy=8 J
in scene B and scene C, as compression =0, spring energy =0
table is filled as below:
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