a constant rrtarding force of50N is applied to a body of mass 20 kg moving initially with speed of 15 ms-1.How long does the body take to stop?
a = F/m = - 50/20 = -2.5
v = Vi + a t
0 = 15 - 2.5 t
t = 15/2.5
Physics
To calculate the time it takes for the body to stop, we need to use Newton's second law of motion and the equation for acceleration.
Newton's second law states that the force acting on an object is equal to its mass multiplied by its acceleration:
F = m * a
In this case, the force exerted on the body is the retarding force, which is equal to 50 N.
The acceleration of the object can be calculated using the formula:
a = F / m
Now, we have the acceleration and the initial velocity of the object. We can calculate the time it takes for the body to stop using the following equation of motion:
v = u + at
where:
v = final velocity (0 m/s, since the body stops)
u = initial velocity (15 m/s)
a = acceleration (calculated from the retarding force)
t = time
Rearranging the equation to solve for time (t), we have:
t = (v - u) / a
Substituting the values into the equation:
t = (0 - 15) / (50 / 20)
First, we need to calculate the acceleration using the formula:
a = F / m
= 50 N / 20 kg
= 2.5 m/s^2
Now, we can find the time it takes for the body to stop:
t = (0 - 15) / 2.5
= -15 / 2.5
= -6
Therefore, the time it takes for the body to stop is 6 seconds. Note that the negative sign indicates that the body is decelerating.