Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson,

*Physics, 2/e*, Problem 6.58, Comprehensive Problem 6.82

An 0.080 kg block traveling horizontally at 5.5 m/s hits a horizontal spring (

*k*= 320 N/m) that is initially uncompressed. Neglect friction and drag. As a result the block is brought to a rest when the spring is compressed by:

(A) 0.037 m.

(B) 0.061 m.

(C) 0.087 m.

(D) 1.2 m.

Correct answer (highlight to unhide): (C)

Starting with the energy balance equation:

*W*= ∆

_{nc}*KE*+ ∆

_{tr}*PE*+ ∆

_{grav}*PE*,

_{elas}where

*W*= 0 (no external gains/losses of mechanical energy), and ∆

_{nc}*PE*= 0 (as there is no change in elevation of the block as it travels horizontally to come to rest), such that:

_{grav}0 = ∆

*KE*+ ∆

_{tr}*PE*,

_{elas}0 = (1/2)·

*m*·∆(

*v*

^{2}) + (1/2)·

*k*·∆(

*x*

^{2}),

0 = (1/2)·

*m*·(

*v*

_{f}^{2}–

*v*

_{0}

^{2}) + (1/2)·

*k*·(

*x*

_{f}^{2}–

*x*

_{0}

^{2}).

With initial parameters of

*v*

_{0}= +5.5 m/s (block is moving) and

*x*

_{0}= 0 (spring is relaxed/uncompressed), and final parameters

*v*= 0 (block at rest), then the distance

_{f}*x*that the spring has been compressed from equilibrium can now be solved for:

_{f}0 = (1/2)·

*m*·(0

^{2}–

*v*

_{0}

^{2}) + (1/2)·

*k*·(

*x*

_{f}^{2}– 0

^{2}),

0 = –(1/2)·

*m*·

*v*

_{0}

^{2}+ (1/2)·

*k*·

*x*

_{f}^{2},

*k*·

*x*

_{f}^{2}=

*m*·

*v*

_{0}

^{2},

*x*=

_{f}*v*

_{0}·(

*m*/

*k*)

^{1/2},

*x*= (

_{f}__5__.

__5__m/s)·((0.080 kg)/(320 N/m))

^{1/2}= 0.08696263565 m,

or to two significant figures, the spring is compressed by 0.087 m.

(Response (A) is (

*m*·

*v*

_{0}/

*k*)

^{1/2}; response (B) is

*v*

_{0}·(

*m*/(2·

*k*))

^{1/2}; response (D) is

*v*

_{0}·(2·

*g*)

^{1/2}.)

Sections 70854, 70855

Exam code: quiz04s7aT

(A) : 12 students

(B) : 16 students

(C) : 17 students

(D) : 3 students

Success level: 35%

Discrimination index (Aubrecht & Aubrecht, 1983): 0.37