Question

A car of mass m=1000 kg starts moving from rest at a point A on a smooth horizontal surface. The car moves under the action of an engine force F ⃗ of magnitude F=2000 N and of direction parallel to the surface. It reaches point O that is 70 m away from A, with a velocity〖 v ⃗〗_f.
Using the Second Law of Newton (F ⃗_net=ma ⃗)
Determine the acceleration of the car.
Calculate the normal force acting on the car.

Calculate the final velocity at O.

Answer 1

Answer:

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Explanation:

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Answer 2

Given:

mass of car = 1000kg

initial velocity(u) = 0, final velocity(v) = v

Engine force(F) = 2000N

distance travelled =  70m

To Find:

acceleration of the car = a =?

Normal force acting on car = N =?

final velocity = v =?

Explanation:

According to the given information, we can draw the situation as shown below,

Now acceleration of car will be

$F = ma$

2000N = 1000kg × a

a = 2m/s²

For Normal force acting on the car,

N = mg

N= 1000 × 10    ( g = 10m/s²)

N = 10000N

For final velocity,

we know that v² - u² = 2as

putting values,

v² = 0 + 2(2)70 = 280

v² = 280

v = $\sqrt{280}$ = 16.73 m/s