The motion of a crate can be divided into two stages. Stage 1 (A-B): The crate with mass m = 4 kg is compressed against a massless spring with a spring constant (stiffness) k = 256 N/m to point A (0.5 m from its equilibrium position O), then released from rest. The crate moves along a horizontal frictionless surface, from A to B, 1 m above the reference level of PEg. Stage 2 (B-C): The crate slides down a rough curved surface until it reaches its bottom at point C with a speed vc = 5 m/s. Calculate the work done on the crate in stage 2 by the force of kinetic friction. Take g = 10 m/s². VA 0 XA h = 1 m = 5 m/s O -22 J O - 11 J -5 J O - 44 J - 33J O -0.5 m PE = 0- Vc

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Chapter8: Potential Energy And Conservation Of Energy
Section: Chapter Questions
Problem 79AP: Consider a block of mass 0.200 kg attached to a spring of spring constant 100 N/m. The block is...
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The motion of a crate can be divided into two stages.
Stage 1 (A-B): The crate with mass m = 4 kg is compressed against a
massless spring with a spring constant (stiffness) k = 256 N/m to point A
(0.5 m from its equilibrium position O), then released from rest. The crate
moves along a horizontal frictionless surface, from A to B, 1 m above the
reference level of PEg.
Stage 2 (B-C): The crate slides down a rough curved surface until it reaches
its bottom at point C with a speed vc = 5 m/s.
Calculate the work done on the crate in stage 2 by the force of kinetic friction.
Take g = 10 m/s².
VA
0
= -0.5 m
h = 1 m
Vc = 5 m/s
ХА
-22 J
O - 11 J
O -5 J
O - 44 J
O-33J
PE
= 0-
Transcribed Image Text:The motion of a crate can be divided into two stages. Stage 1 (A-B): The crate with mass m = 4 kg is compressed against a massless spring with a spring constant (stiffness) k = 256 N/m to point A (0.5 m from its equilibrium position O), then released from rest. The crate moves along a horizontal frictionless surface, from A to B, 1 m above the reference level of PEg. Stage 2 (B-C): The crate slides down a rough curved surface until it reaches its bottom at point C with a speed vc = 5 m/s. Calculate the work done on the crate in stage 2 by the force of kinetic friction. Take g = 10 m/s². VA 0 = -0.5 m h = 1 m Vc = 5 m/s ХА -22 J O - 11 J O -5 J O - 44 J O-33J PE = 0-
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