A 110 g block attached to a spring with spring constant 2.1 N/m oscillates horiz

Question

A 110 g block attached to a spring with spring constant 2.1 N/m oscillates horizontally on a frictionless table. Its velocity is 20 cm/s when x0 = -4.8 cm .

Part A

What is the amplitude of oscillation?

Express your answer to two significant figures and include the appropriate units.

Part B

What is the block's maximum acceleration?

Express your answer to two significant figures and include the appropriate units.

Part C

What is the block's position when the acceleration is maximum?

Express your answer to two significant figures and include the appropriate units.

Part D

What is the speed of the block when x1 = 3.1 cm ?

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Solution:Part A

At 4.6cm the ball's energy = potential energy + kinetic energy
1/2 kXo^2 + 1/2 mv^2

At the amplitude the velocity of the ball is zero, so it's energy = potential energy
= 1/2 kA^2

presuming no energy loss,
1/2 kXo^2 + 1/2 mv^2 = 1/2 kA^2
( 1/2 x 2.1 x 0.048^2 ) + ( 1/2 x 0.10 x 0.20^2 ) = 1/2 x 2.1 x A^2
2.4192 x 10^-3 + 2 x 10^-3 = 1.05 A^2
A = 0.0649 m
= 6.49 cm

Part B
Maximum force on the ball is when the ball is at the amplitude. So the maximum acceleration is also when the ball is at the amplitude. The force on the ball,
F = kx
The force makes the ball accelerate towards the center,
F = ma
ma = kx
0.10 x a = 2.1 x 0.0649
a = 1.3629 ms^-2

Part C

What is the block's position when the acceleration is maximum?

= 6.49 cm

Part D
1/2 kXo^2 + 1/2 mVo^2 = 1/2 kX1^2 + 1/2 mV1^2

( 2.1 x 0.048^2 ) + ( 0.10 x 0.20^2 ) = ( 2.1 x 0.031^2 ) + ( 0.10 x V1^2 )
V1 = 0.26 ms^-1
= 26 cms^-1

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