A 28 g bullet pierces a sand bag 30 cmthick. If the initial bullet velocity was 55m/s and it emerged from the sandbag with18 m/s what is the magnitude of the frictionforce (assuming it to be constant) the bullet experienced while ittraveled through the bag?
a. 130 N
b. 1.3 N
c. 13 N
d. 38 N

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 kg

Thickness of sand bag m

Initial bullet velocity

Final bullet velocity

According to the work energy theorem,

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

Here we have to find friction force

  N

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

Therefore, the frictional force is 126.046 N

answera compound;

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft