A compressed spring is placed between carts of mass 1 kg and 3 kg, and both carts are held in place by catches. The catches are released simultaneously.
Adjust conditions:
The sliders adjust the values
放した直後、1 kgの車の加速度の大きさは3 kgの車の何倍か？
摩擦係数0.10で放した直後、軽い車の加速度の大きさは重い車の何倍か？
The equations explain the phenomenon
1How velocity changesa=dvdt……
2Measuring force with a stretched or compressed springF=kx……
3Comparing mass and accelerationma=F……
4Expressing acceleration as a derivativemdvdt=F……
5Including gravity on the right-hand sidemdvdt=mg……
6The acceleration of a stationary boxF=mdvdt=0……
7The force that balances gravityN−mg=0……
8Table compression and the normal forceN=kΔx……
9The origins of contact forcesmdvdt=Fgravity+Felectromagnetic……
10The forces the box and table exert on each otherFbox→table=−Ftable→box……
11The equation of motion for each cart|a1||a2|=m2m1……
12Include friction and write the equations of motion for both carts|a1||a2|=(m2m1)×(F − μm1g)/(F − μm2g)……
What this chapter has shown
The right-hand side of the equation of motion is the sum of all forces acting on the chosen object, including their directions. For the same force magnitude, a smaller mass has a greater acceleration magnitude. The forces discussed in this chapter originate from gravity and electromagnetic interactions, and the forces paired by Newton's third law act on different objects.
3 timesRatio of acceleration magnitudes
With the wheels locked and the coefficient of friction set to 0.10, the ratio immediately after release is 5.
In everyday lifeWhile 2 people push against each other on ice, they experience forces of equal magnitude, and the lighter person has the greater acceleration magnitude.
The equation of motion m dv/dt = F relates the rate of change of the box's velocity to the net force acting on it. The forces on the right-hand side can be classified as gravity and electromagnetic forces. The table also compressed slightly, supporting the box through electromagnetic forces between atoms. Understanding which object exerts a force on which other object, and identifying the corresponding force pair, makes it clear why we set up the equation as we do, beyond simply memorizing the formula.