Question

a particle of mass 50g moving on a straight line . the variation speed with time is shown in the figure. find the force acting the particle at t=2,4,6 sec​

Answer 1

Answer:

0.25N opposite to the motion

Explanation:

m=50g=5×10

−2

kg

As shown in the figure.

Slope of OA=tanθ

OD

AD

=

3

15

=5 m/s

2

So, at t=2 sec acceleration is 5 m/s

2

Force =ma=5×10

−2

×5=0.25 N along the motion.

At t=4sec

slope of AB=0, acceleration =0[tan0

o

=0]

∴ Force =0

At t=6 sec, acceleration = slope of BC

In ΔBEC=tanθ=

EC

BE

=

5

15

=5.

Slope of BC=tan(180

o

−θ)=−tanθ=−5 m/s

2

(deceleration)

Force =ma=5×10

−2

5=0.25 N. Opposite to the motion.