Block P of mass 2 kg slides down the surface and has a speed 20 m/s at the lowest point, Q, where the local radius of curvature is 2 m as shown in the figure. Assuming g = 10 m/s² , the normal force (in N) at Q is _______ (correct to two decimal places).

The correct answer is 420.

Based on the provided diagram, block P of mass $m$ slides down a curved surface and passes through its lowest point Q, where the local radius of curvature is $R$. We are given the following values from the problem statement:

Mass of the block, $m=2\text{kg}$
Speed of the block at Q, $v=20\text{m/s}$
Local radius of curvature at Q, $R=2\text{m}$
Acceleration due to gravity, $g=10{\text{m/s}}^2$

At the lowest point Q, the forces acting on the block in the vertical direction are:

1. The gravitational force (weight) acting vertically downwards:
$W=mg$

2. The normal force $N$ exerted by the track on the block, acting vertically upwards (towards the center of curvature of the path).

The net vertical force directed towards the center of curvature provides the centripetal acceleration required for the circular path at this point:

$N-mg=\frac{m{v}^2}{R}$

We can solve for the normal force $N$ by rearranging the terms:

$N=mg+\frac{m{v}^2}{R}$

Substitute the given values into the equation:

$N=\left(2\text{kg}\right)\left(10{\text{m/s}}^2\right)+\frac{\left(2\text{kg}\right){\left(20\text{m/s}\right)}^2}{2\text{m}}$

Simplify the expression:

$N=20+\frac{2\times 400}{2}$

$N=20+400$

$N=420\text{N}$

Thus, the normal force acting on the block at point Q is 420 N.