Question

A 5.00 kg projectile has a velocity of 255 m/s right. Find the force to stop this projectile in 1.45 s.

Answer 1

Answer:

-879.3N

Explanation:

F=ma

a=u-v/t

=0-255/1.45

=-255/1.45

a=-175.8

F=ma

=5(-175.7

F=-879.3N

Answer 2

Answer: The force which is required to stop a 5 kg projectile is  

F= -0.879

Explanation:

Projectile is an object which thrown into space on which the only source of acting force is gravity. The path followed by the projectile is called trajectory. Acceleration is in the horizontal motion and vertical projectile motion of a particle.

Acoording to the law of conservation of linear momentum,

if the net external force acting on the body is zero, then the momentum of that body remains constant. The momentum of the body in the system can increase or decrease as per the situation but the momentum of the system is always conserved.

According to Newton's second law of linear momentum , the net external force is equal to the change in the momentum of a system divided by the time in which it changes.

Given data: m= 5.10⁻³kg v₀=255m/s₀, Δt=1.45

According to the law of conversation of linear momentum, change in linear momentum can be expressed by

Δp=mv-mv₀

   m(v-v₀)

where m= mass, v= final speed, v₀=initial speed

Also, we can assume that according to the Newton's second law linear momentum is given by:

According to Newton's second law of linear momentum is expressed as

p=ΔtF

where p= linear momentum, Δt= time interval, F= force

Equating above two highlighted equations we get

m(v-v₀)=ΔtF

So we can express force from this equation as

F= m(v-v₀)/ΔtF

Putting values in the above equation

F= 5.10⁻³ x (0-255)/1.45

  = -0.879N

Because of the negative sign present  in the value we can say that force is in opposite direction of the motion.

Thus the force required to stop the projectile is F= -0.879

To learn more about Physics, click on the link below

https://brainly.in/question/434606

https://brainly.in/question/93876

#SPJ3