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  1. Problems
  2. /Two blocks, of mass 4.0 kg and 2.5 kg, touch each other on a smooth floor…
MediumPhysicsHigh School (11–12)Mechanics › Newton's Laws of MotionNumericalNEETBoard ExamAPA-LevelIB

Two blocks, of mass 4.0 kg and 2.5 kg, touch each other on a smooth floor…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Two blocks, of mass 4.0 kg4.0\ \text{kg}4.0 kg and 2.5 kg2.5\ \text{kg}2.5 kg, touch each other on a smooth floor. A horizontal push of 13 N13\ \text{N}13 N acts on the 4.0 kg4.0\ \text{kg}4.0 kg block, which pushes the other.

  1. What is their acceleration?
  2. What force does each block exert on the other?
  3. What would this force be if the same push acted on the 2.5 kg2.5\ \text{kg}2.5 kg block instead?

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

m/s²
N
N
Need a reminder? The key ideaShowHide

Bodies joined by a light, taut string (or pushing on each other) move with the same acceleration. First treat the whole system: a=net outside forcetotal massa = \dfrac{\text{net outside force}}{\text{total mass}}a=total massnet outside force​. Then take one body alone to find the force between them.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. m1=4.0 kgm_1 = 4.0\ \text{kg}m1​=4.0 kg (pushed), m2=2.5 kgm_2 = 2.5\ \text{kg}m2​=2.5 kg, F=13 NF = 13\ \text{N}F=13 N.

Plan. System for aaa, then the front block alone for the contact force.

Step 1. Acceleration. a=13 N4.0 kg+2.5 kg=2.0 m/s2a = \dfrac{13\ \text{N}}{4.0\ \text{kg} + 2.5\ \text{kg}} = 2.0\ \text{m/s}^2a=4.0 kg+2.5 kg13 N​=2.0 m/s2.

Step 2. The front block. Only the contact force pushes it: Fc=2.5 kg×2.0 m/s2=5.0 NF_c = 2.5\ \text{kg} \times 2.0\ \text{m/s}^2 = 5.0\ \text{N}Fc​=2.5 kg×2.0 m/s2=5.0 N.

Step 3. Pushing the other side. Now the 4.0 kg4.0\ \text{kg}4.0 kg block is in front: Fc=4.0 kg×2.0 m/s2=8.0 NF_c = 4.0\ \text{kg} \times 2.0\ \text{m/s}^2 = 8.0\ \text{N}Fc​=4.0 kg×2.0 m/s2=8.0 N.

Answer: (a) 2.0 m/s22.0\ \text{m/s}^22.0 m/s2. (b) 5.0 N5.0\ \text{N}5.0 N each way (third law). (c) 8.0 N8.0\ \text{N}8.0 N.

Check. The pushed block: 13 N−5.0 N=8.0 N13\ \text{N} - 5.0\ \text{N} = 8.0\ \text{N}13 N−5.0 N=8.0 N, and 4.0 kg×2.0 m/s2=8.0 N4.0\ \text{kg} \times 2.0\ \text{m/s}^2 = 8.0\ \text{N}4.0 kg×2.0 m/s2=8.0 N. It agrees.

Watch out. The contact force is not the whole push. Part of the push speeds up the first block itself.

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