Question

A constant force at on an object of mass height kg for duration of 2 second it increses the onject velocity from 3 m per second to 7m per second find the magnitude of applied force now if a force was applied force duration 5 second what would final velocity of an object

Answer 1

Answer:

t1 = 5s, t2 = 2 , v = 7m/s ,

u = 3m/s

let the mass of object be x kg

F = m a

= x ×( v -u/t1 - t2)

= x ×( 7 -3/5- 2)

= x × (5/3)

= 1.66x N

To find final velocity if, duration of applied force was 5 second.

note:( t1-t2= 5-2 = 3 )

then , 1.66x = x × ( v - u/3)

1.66 × 3 = v - u

4.98 = v- 3

4 . 98 + 3 = v

v= 7.98m/s

Approximate, v = 8m/s

magnitude of force is 1.66x newton.

And final velocity (v)= 8m/s