If dot c moves to position 5 (60 meters), what is - id560410386, elainestorm9633

Question

Physics: If dot c moves to position 5 (60 meters), what is its average velocity during these 5 seconds? a) 10 m/s b) 11 m/s c) 12 m/s d) 13 m/s

elainestorm9633 · Accepted Answer

Here not any option is matching but only (A) option is near to our answer.

The frictional force is 126.046 N

Explanation:

Given:

Mass of bullet $m = 28 \times 10^{-3}$ kg

Thickness of sand bag $d = 30 \times 10^{-2}$ m

Initial bullet velocity $v_{i} = 55 \frac{m}{s}$

Final bullet velocity $v_{f} = 18 \frac{m}{s}$

According to the work energy theorem,

Work done by friction force is equal to change in kinetic energy.

$\frac{1}{2} m v_{i} ^{2} - \frac{1}{2} m v_{f } ^{2} = f_{s} d$

Here we have to find friction force $f_{s}$

$\frac{1}{2} \times 0.028 (3025-324) = f_{s} \times 0.30$

$f_{s} = 126.046$ N

Here not any option is matching but only (A) option is near to our answer.

Therefore, the frictional force is 126.046 N

If dot c moves to position 5 (60 meters), what is its average velocity during these 5 seconds? a) 10 m/s b) 11 m/s c) 12 m/s d) 13 m/s