A 0.297-kg block resting on a frictionless, horizontal surface is attached to a
ID: 1448715 • Letter: A
Question
A 0.297-kg block resting on a frictionless, horizontal surface is attached to a spring having force constant 83.8 N/m as in the figure below. A horizontal force
causes the spring to stretch at a distance of 5.69 cm from its equilibrium position.
(a) Find the value of F. (Enter the magnitude of the force only.)
N
(b) What is the total energy stored in the system when the spring is stretched?
J
(c) Find the magnitude of the acceleration of the block immediately after the applied force is removed.
m/s2
(d) Find the speed of the block when it first reaches the equilibrium position.
m/s
(e) If the surface is not frictionless but the block still reaches the equilibrium position, how would your answer to part (d) change?
The block would arrive at a greater speed.
The block would arrive at a lower speed.
The block would arrive at the same speed.
Explanation / Answer
a) F=kx = 83.8*0.0569 = 4.77 N
b) ETotal = 1/2kA^2 = ½*83.8*0.0569^2 = 0.136 J
c) Fnet = ma
-kx = ma
a= (-kx)/m = (-83.8*0.0569)/0.297 = -16.05 m/s^2
d) By law of conservation of energy
KE= ETotal
1/2mv^2 = ETotal
½*0.297*v^2 = 0.136 => v= 0.97m/s
e)
The block would arrive at the same speed.
Reason: By the law of conservation of energy there is no loss in gravitational or in frictional forces.
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