Question

The average resisting force that must act on a 5 kg mass to reduce its speed from 65 cm/s to 15 cm/s in 0.2 s is:

• A. 12.5 N
• B. 25 N
• C. 50 N
• D. 100 N

Answer 1

The correct option is A 12.5 N

$Fdt=mv-mu$ from work energy theorem

$F=\frac{m(v-u)}{dt}=\frac{5(65-15)}{0.2\times 100}=\frac{5\times 50}{20}=12.5N$