Question

06. A foree of 4 N acts on a body of mass 2 kg for 4 s. assuming
Time TS
the body to be initially at rest, find (a) its velocity when the force stops acting (b) the
distance čovered in 10 s after the force starts acting.
​

Answer 1

Answer:

a) 8m/s b) 64m

Explanation:

Force=mass × acceleration

Thus,

4=2×a

Therefore a=2 m/s^2

Now answer to (a),

We know that,

v=u+at (v is final velocity and u is initial velocity)

v=0+(2×4)

=8m/s [Since no frictional force is given we assume body to be moving on frictionless surface]

Now the answer to (b),

We know that,

Distance = ut + at^2/2

Distance = (0*4)+(2*16/2)

=16m for the first 4s, but after that any opposing force is not given.

Assuming that the ball moves on frictionless surface, here u becomes 8m/s and acceleration becomes zero.

Time left is 6s.

Therefore,

Distance =(8*6)+(0*6/2) [ut +at^2/2]

=48m

Thus total distance covered is 48 + 16 = 64m.

Answer 2

see, first we should understand, what is the area of force-time graphs shows. It's the impulse. And impulse is equals to change in momentum.

first find impulse,

impluse = area

            =((1/2)×2×10)+ (2×10)+ ((1/2)×(10+20)×2)+((1/2)×4×20)

            =10+20+30+40

            =100 N-s

now, impulse = change in momentum

                     =m×v - m×u

where v= final velocity and u =initial velocity =0

so, 100 = 2×v  

⇒v = 50 m/s

hope this will help you

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