Question 1

Multiple choiceHard1 mark

Blocks A, B and C, of masses 2.0 kg, 3.0 kg and 5.0 kg respectively, are in contact in that order on a horizontal track. A 50 N horizontal force pushes B to the right. Friction forces on A, B and C are 2.0 N, 5.0 N and 3.0 N respectively, all to the left. The blocks move together. What is the force exerted by C on B?

Show answer and explanation

Answer: B. For all three blocks, the resultant force is 50−2.0−5.0−3.0=40 N50-2.0-5.0-3.0=40\ \text{N} so their acceleration is 40/(2.0+3.0+5.0)=4.0 m s−240/(2.0+3.0+5.0)=4.0\ \text{m s}^{-2} to the right. For C, the contact force from B satisfies F−3.0=5.0(4.0)F-3.0=5.0(4.0) giving F=23 NF=23\ \text{N} to the right. The force exerted by C on B is equal in magnitude and opposite in direction: 23 N23\ \text{N} to the left.
A: uses mCa=20 Nm_Ca=20\ \text{N} as the contact force, omitting friction on C.
C: gives the force on C by B, not the equal-and-opposite force on B.
D: treats the resultant force on all three blocks as the force at the B–C contact.
Remember: interaction forces act on different bodies and are equal in magnitude and opposite in direction.

More A.2 Forces and momentum questions