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2 years ago

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A 1500 kg car is pushing a 4000 kg truck. The car and truck are accelerating at 2.0 m/s^2. Assuming that the frictional force on

the truck is negligible, what force is the truck exerting on the car?

1 answer:

Crazy boy [7]2 years ago

7 0

To solve this problem we will use the Force equation according to the definition given in Newton's second law. There we have that the Force is equal to

$F =ma$

Where,

m = mass

a = acceleration

Our values are given as

$m_1 = 1500 kg$

$m_2 = 4000 kg$

$a = 2.0 m/s^2$

Considering that both mass are equal to one, we have that:

$F = (m_1+m_2)* a$

$F = (1500 + 4000)(2.0)$

$F = 11000 N = 11kN$

Therefore the truck exert a force on the car of 11kN

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Maximum height reached = 35.15 meter.

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In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

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In the give problem we have R = 301.5 m, θ = 25° we need to find H.

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