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Physics / Chapter 5: Force and Motion - I / Section 5-4

Mass — Question 10

Section 5-4 · Physics pages 690–691

Question 10 A hidden payload changes the acceleration calibration An empty cart accelerates at 5 m/s2 under a fixed 20 N net force. With a payload, the same net force produces 2 m/s2. Both cart and payload move together. cart F = 20 N, a = 5 m/s2 empty cart F = 20 N, a = 2 m/s2 loaded load Schematic; labels specify the data. Arrow lengths are not a common scale. (a) Find the empty and loaded masses. (b) Find the payload mass. (c) If the loaded acceleration were 10 m/s2 instead, would the data describe adding a positive payload under the same net force?
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Question 10 - Solution A hidden payload changes the acceleration calibration cart F = 20 N, a = 5 m/s2 empty cart F = 20 N, a = 2 m/s2 loaded load (a) Infer both total inertial masses. mE = 20/5 = 4 kg, mL = 20/2 = 10 kg The empty mass is 20/5 = 4 kg. The loaded mass is 20/2 = 10 kg. (b) Subtract the unchanged cart. mP = mL −mE = 10 −4 = 6 kg The payload mass is 10 −4 = 6 kg. The force acts on the total moving mass. (c) Test physical consistency. m′ L = 20/10 = 2 kg ⇒m′ P = 2 −4 = −2 kg < 0 An acceleration of 10 m/s2 would imply only 2 kg total, less than the empty 4 kg. A positive added payload cannot explain that under the same net force. At least one stated assumption or measurement would need revision.

Question 10

A hidden payload changes the acceleration calibration

An empty cart accelerates at 5 m/s2^2 under a fixed 20 N net force. With a payload, the same net force produces 2 m/s2^2. Both cart and payload move together.

Diagram shown in the original SVG worksheet.

  1. Find the empty and loaded masses.

  2. Find the payload mass.

  3. If the loaded acceleration were 10 m/s2^2 instead, would the data describe adding a positive payload under the same net force?

Question 10 - Solution

A hidden payload changes the acceleration calibration

Diagram shown in the original SVG worksheet.

(a) Infer both total inertial masses.

mE=20/5=4kg,mL=20/2=10kgm_E=20/5=4\,\mathrm{kg},\quad m_L=20/2=10\,\mathrm{kg} The empty mass is 20/5=420/5=4 kg. The loaded mass is 20/2=1020/2=10 kg.

(b) Subtract the unchanged cart.

mP=mL−mE=10−4=6kg\boxed{m_P=m_L-m_E=10-4=6\,\mathrm{kg}} The payload mass is 10−4=610-4=6 kg. The force acts on the total moving mass.

(c) Test physical consistency.

mL′=20/10=2kg⇒mP′=2−4=−2kg<0m_L'=20/10=2\,\mathrm{kg}\ \Rightarrow\ m_P'=2-4=-2\,\mathrm{kg}<0 An acceleration of 10 m/s2^2 would imply only 2 kg total, less than the empty 4 kg. A positive added payload cannot explain that under the same net force. At least one stated assumption or measurement would need revision.

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