Question

A force of 200N acts on a body of mass 20kg initially rest for 5 seconds. Determine i) acceleration ii) final velocity attained and iii) distance covered by it

Answer 1

Answer:

F=200N

M=20kg

1. t=5SF=MA

A=M/F

A=200/20=10m/s^-2

2)since

u=0

v=u+At

v=10*5=50s

3)s=ut+A*t^2/2

S=10*100/2

s=500m

s=distance

t=time

A=acceleration

u=initial velocity

v=final velocity

F=force

Answer 2

Explanation:

Given that,

Force acting on the object, F = 200 N

Mass of the object, m = 20 kg

Initially it is at rest, u = 0

1. Force acting on the object is given by Newton’s second law of motion as :

$a=\dfrac{F}{m}$

$a=\dfrac{200}{20}=10\ m/s^2$

2. Let v is the final speed of the object. It is given by :

v = u + at

v = 0 + 10 × 5

v = 50 m/s

3. Let s is the distance covered by the object. Using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

$s=\dfrac{1}{2}\times 10\times (5)^2$

s = 75 m

Hence, this is the required solution.