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Answer :
To solve the problem, we start with Newton's second law of motion:
[tex]$$
F = m \cdot a
$$[/tex]
where:
- [tex]$F$[/tex] is the force,
- [tex]$m$[/tex] is the mass,
- [tex]$a$[/tex] is the acceleration.
Step 1: Calculate the force on the 64 kg block
For the block with mass [tex]$64\ \text{kg}$[/tex] and acceleration [tex]$2\ \text{m/s}^2$[/tex], the force is:
[tex]$$
F_{64} = 64\ \text{kg} \times 2\ \frac{\text{m}}{\text{s}^2} = 128\ \text{N}
$$[/tex]
Step 2: Calculate the force on the 60 kg block
For the block with mass [tex]$60\ \text{kg}$[/tex] and the same acceleration [tex]$2\ \text{m/s}^2$[/tex], the force is:
[tex]$$
F_{60} = 60\ \text{kg} \times 2\ \frac{\text{m}}{\text{s}^2} = 120\ \text{N}
$$[/tex]
Step 3: Compare the forces
- The force on the [tex]$64\ \text{kg}$[/tex] block is [tex]$128\ \text{N}$[/tex].
- The force on the [tex]$60\ \text{kg}$[/tex] block is [tex]$120\ \text{N}$[/tex].
Since
[tex]$$
120\ \text{N} < 128\ \text{N},
$$[/tex]
the force acting on the [tex]$60\ \text{kg}$[/tex] block is smaller.
Final Answer: The force on the [tex]$60\ \text{kg}$[/tex] block is smaller.
[tex]$$
F = m \cdot a
$$[/tex]
where:
- [tex]$F$[/tex] is the force,
- [tex]$m$[/tex] is the mass,
- [tex]$a$[/tex] is the acceleration.
Step 1: Calculate the force on the 64 kg block
For the block with mass [tex]$64\ \text{kg}$[/tex] and acceleration [tex]$2\ \text{m/s}^2$[/tex], the force is:
[tex]$$
F_{64} = 64\ \text{kg} \times 2\ \frac{\text{m}}{\text{s}^2} = 128\ \text{N}
$$[/tex]
Step 2: Calculate the force on the 60 kg block
For the block with mass [tex]$60\ \text{kg}$[/tex] and the same acceleration [tex]$2\ \text{m/s}^2$[/tex], the force is:
[tex]$$
F_{60} = 60\ \text{kg} \times 2\ \frac{\text{m}}{\text{s}^2} = 120\ \text{N}
$$[/tex]
Step 3: Compare the forces
- The force on the [tex]$64\ \text{kg}$[/tex] block is [tex]$128\ \text{N}$[/tex].
- The force on the [tex]$60\ \text{kg}$[/tex] block is [tex]$120\ \text{N}$[/tex].
Since
[tex]$$
120\ \text{N} < 128\ \text{N},
$$[/tex]
the force acting on the [tex]$60\ \text{kg}$[/tex] block is smaller.
Final Answer: The force on the [tex]$60\ \text{kg}$[/tex] block is smaller.
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