Question

The speed of a vehicle is reduced with a constant acceleration from 72km/h to 18

km/h over 250m directly down an incline of lin25. The mass of the vehicle is 1.9
Mg and it has a constant resistance to motion of 350N.
Calculate the magnitude of the braking Force
Answer:
Check​

Answer 1

The correct answer will be "1477.84 N".

Step-by-step explanation:

Given that,

Mass,

m = 1.6 mg

or,

= 1600 kg

Initial velocity,

u = 72 km/h

=
`72* (5)/(18) \ m/s`

=
`20 \ m/s`

Final velocity,

v = 18 km/h

=
`18* (5)/(18)`

=
`5 \ m/s`

Covered distance,

s = 250 m

By using the below relation, we get

⇒
`v^2=u^2+2as`

On putting the values, we get

⇒
`(5)^2=(20)^2+2* a* 250`

⇒
`a=-0.75 \ m/s^2` (shows the deceleration)

Slope will be given as 1 in 25, then

⇒
`Sin \theta=(1)/(25)`

`\theta=2.3^(\circ)`

hence,

As we know,

⇒
`\Sigma F=ma`

or,

⇒
`Braking \ force+350-mgSin\theta=ma`

⇒
`Braking \ force=ma+mgSin\theta-350`

On substituting all the values, we get

⇒
`=1600(0.75+1600* 9.81 Sin(2.3^(\circ))-350`

⇒
`=1477.84 \ N`