# Determine the magnitude of the maximum acceleration

## Homework Statement

A block of mass m = 0.677 kg is fastened to an unstrained horizontal spring whose spring constant is k = 88.6 N/m. The block is given a displacement of +0.170 m, where the + sign indicates that the displacement is along the +x axis, and then released from rest. (a) What is the force (with sign) that the spring exerts on the block just before the block is released? (b) Find the angular frequency of the resulting oscillatory motion. (c) What is the maximum speed of the block? (d) Determine the magnitude of the maximum acceleration of the block.

## Homework Equations

(a) F(applied) = kx
(b) angular freq = sq rt(k/m)
(c) (1/2)kx^2 = (1/2) mv^2
(d)a = m/(kx)

## The Attempt at a Solution

(a) F = 88.6 n/m * 0.170 m = +15.062 N

(b) I got 11.4399 rad/s which was correct

(c) I got 1.944 m/s which was correct

(d) a = 0.677 kg / (88.6 N/m * 0.170 m) = 0.044948

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berkeman
Mentor
Looks good. For (d), you could have just written a = F/m, and used your answer for max F from part (a).

alphysicist
Homework Helper
Hi Kris1120,

I think a couple of your relevant equations are incorrect, and is giving your wrong answers to part a and d.

## Homework Statement

A block of mass m = 0.677 kg is fastened to an unstrained horizontal spring whose spring constant is k = 88.6 N/m. The block is given a displacement of +0.170 m, where the + sign indicates that the displacement is along the +x axis, and then released from rest. (a) What is the force (with sign) that the spring exerts on the block just before the block is released? (b) Find the angular frequency of the resulting oscillatory motion. (c) What is the maximum speed of the block? (d) Determine the magnitude of the maximum acceleration of the block.

## Homework Equations

(a) F(applied) = kx
This is true if all they want is the magnitude; but here they want the sign. The correct formula (in terms of the vector F and vector x) is:

$$\vec F = - k \vec x$$

(b) angular freq = sq rt(k/m)
(c) (1/2)kx^2 = (1/2) mv^2
(d)a = m/(kx)
This equation is not correct. You need to start by setting the expressions for the force magnitudes together, which is

$$m a = k x$$

and then solve for a. What do you get for an answer?

berkeman
Mentor
Yikes, that will teach me to pop in and try to help. Thanks for the corrections alphysicist.

Ok so on part (a) it should be negative and on part (d) my equation was upside down! Thank you so much for your help!