Question

Find the value of the constant force that is applied for 1 second on a 10kg object to increase the velocity of the object from 5 m/s to 10 m/s.​

Answer 1

Explanation:

F=dp/dt

F÷m(v-u)/t

F=10(10‐5)/1

F= 50N

Answer 2

Answer:

The constant force that is applied to the object is 50N.

Explanation:

The force is defined as the rate of change of momentum per unit of time. i.e.

$F=\frac{P}{t}$        (1)

Where,

F=force acting on the object

P=momentum of the object

t=time for which the momentum changes

From the question we have,

t=1 second

Mass of the object(m)=10kg

Initial velocity(v₁) of the object=5m/s

Final velocity(v₂) of the object=10m/s

The momentum of an object is given as,

$P=m(v_2-v_1)$          (2)

By substituting the values in equation (2) we get;

$P=10(10-5)=10\times 5=50kgm/s$  (3)

By putting the value of momentum and time in equation (1) we get;

$F=\frac{50}{1}=50N$

Hence, the constant force that is applied to the object is 50N.