a force is applied to a body of mass 0.9 kgs that is at rest the force is applied for a duration of 5 sec and as a result the body covers a distance of 250 m find the magnitude of force
since the force applied gives the body kinetic energy from rest:fd=1/2mv^2 d=1/2at^2 250=1/2a(5)(5) a=20m/s^2 since body start from rest u=0:a=v/t 20=v/5 v=100m/s from the equation:f(250)=1/2*0.9*100^2 250F=4500 F=18N
To find the magnitude of the force applied to the body, we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) of an object multiplied by its acceleration (a): F = m * a.
However, we are given the mass (m) of the body and not the acceleration directly. To find the acceleration, we can use the kinematic equation: s = ut + (1/2)at^2, where s is the distance traveled, u is the initial velocity (which is zero as the body is at rest), t is the time, and a is the acceleration.
Given:
Mass of the body (m) = 0.9 kg
Distance traveled (s) = 250 m
Time (t) = 5 s
First, let's find the acceleration:
s = ut + (1/2)at^2
250 = 0 + (1/2)a(5)^2
250 = (1/2) * 25a
250 = 12.5a
a = 250 / 12.5
a = 20 m/s^2
Now, we can calculate the magnitude of the force:
F = m * a
F = 0.9 kg * 20 m/s^2
F = 18 N
Therefore, the magnitude of the force applied to the body is 18 Newtons.