Question

A particle of mass 50 g moves in a straight line. The variation of speed with time is shown in figure (5−E1). Find the force acting on the particle at t = 2, 4 and 6 seconds.

Answer 1

Answer:

F = 0.25 N opposite the motion.

Explanation:

Given :

mass =50g = 50 x 10 ⁻² kg

Slope of the v-t graph gives acceleration.

At t = 2 s,

Slope =

15/3=5 m/s²

So, acceleration, a = 5 m/s²

F = ma = 5 × 10⁻² × 5

⇒ F = 0.25 N along the motion.

At t = 4 s,

Slope = 0

So, acceleration, a = 0

⇒ F = 0

At t = 6 sec,

Slope =

-15/3= -5 m/s²

So, acceleration, a = − 5 m/s²

F = ma = − 5 × 10⁻² × 5

⇒ F = − 0.25 N along the motion

or, F = 0.25 N opposite the motion.

Answer 2

Answer:

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.