Question

1000 kg vehicle moving with a speed of 20 metre per second is brought to rest at a distance of 50 metre. calculate the unbalanced force acting on the vehicle.​

Answer 1

Explanation:

Let's first calculate Acceleration using stopping distance formula

S= U^2/2a

a= U^2/2S

a= 20×20/ 2× 50

a= 200/50= 4m/s^2

F= Ma

F= 1000× 4

F= 4000N

This is unbalanced force because The force acts on opposite direction and no force is applied in motion direction because it is tending to stop and if Friction is also present in opposite direction

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Answer 2

Answer:

A 1000 kg vehicle moving with a speed of 20 m/s is

brought to rest in a distance of 50 metres :

(1) Find the acceleration.

(2) Calculate the unbalanced force acting on the vehicle.

☆☆Answer ☆☆

$\mathbf \red{given}$

M=1000 kg(mass)

u=20m/s(initial position)

v=0m/s(final position)

s=50m(distance)

♡using newton's second equation of motion

(i)V²=u²+2as

0²=(20)²+2×a×50

0²=400+50a

-400=50a

a=-400/50

a=-4

(ii) we know that unbalanced force equal to mass × acceleration .

so, f=m×a

=>f=1000×-4

=>f=-4000N

Thanks