Calculate resistance force of pushing a car.

Question

the question is as follows;

A car of mass $1000kg$ can be pushed along with an acceleration of $0.2ms^{-2}$ by one man. When a second man exerting the same force comes to his aid, the acceleration is $0.5ms^{-2}$.

There is a constant resistance force $R$ acting horizontally on the car. Calculate $R$.

My attempt;

The force exerted by the first man is $1000kg(0.2ms^{-2})=200kgms^{-2}=200N$. As the second man exerts the same force ($200N$), the total force exerted by both men is $400N$.

For the car to move at a rate of $0.5ms^{-2}$, $1000kg(0.5ms^{-2})=500N$ of force should be required?

Here is where I'm stuck as it seems to me that I need an extra $100N$ to accelerate the car to the required acceleration so the resistance force must be $R=400N-500N=-100N$??

I'm really having a hard time visualising what is going wrong here.

Thanks.

Answer 1

One man pushing ...

$$ F_{net}= F_a - R = 1000(0.2)=200 \tag1$$

Two men push $$ F_{net}= 2F_a - R = 1000(0.5)=500 \tag2$$

eliminate $F_a$

$$ 2*(1) - (2) \implies -R = -100 \implies R=100N$$