Question

What net force is required to bring a 1500 kg car to rest from a speed of 100 km/h over a distance of 55 m?

Answer 1

This is one of the KINEMATIC EQUATIONS applied, for once, to DECLERATION, and INCLUDING a force. So, it ALSO "wants" F=ma, Newton's second law of motion, albeit in formula form (N2 goes on about rates of change of MOMENTUM)
v squared = u squared minus 2 Force/mass s.
v squared = v = 0.
0 =[100x10^3/(60x60)]^2-Force/1500 x 55
If my scribbling is right, solve for Force to get ... force.
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Answer 2

F = ma
Mass is given. Acceleration is not, but you can solve for it using one of the kinematic equations, you can use the: \[v_{f}^{2} = v_{i}^{2} + 2ad\]

(final velocity)^2 = (initial velocity)^2 + 2(acceleration)(distance)

final velocity is 100 km/h. Initial velocity is 0 km/h because of the word "rest." Distance is 55 m.

You can derive the acceleration from the formula \[v_{f}^{2} = 2ad\]

\[a = \frac{ v_{f}^{2} }{ 2d } = \frac{ 10 000 \frac{ km }{ h } }{ 110 } = ?\]

Multiply the acceleration by the mass (1 500 kg) to get the force.

Answer 3

It must be:
\[a = \frac{ 10 000 \frac{ km^ {2} }{ h^ {2}} }{ 110 m }\] not
\[a = \frac{ 10 000 \frac{ km }{ h} }{ 110 }\]

You can convert first 110 m to km.
\[110 m \times \frac{ 1 km }{ 1 000 m } = ?\]
and then divide it to 10 000 km^2/h^2.

Answer 4

From that, you can now find the net force.