Two children are pulling and pushing a 30.0 kg sled. the child pulling the sled is exerting a force of 12.0 n at a 45o angle. the child pushing the sled is exerting a horizontal a force of 8.00 n. there is a force of friction of 5.00 n. what is the weight of the sled? n what is the normal force exerted on the sled? round the answer to the nearest whole number. n what is the acceleration of the sled? round the answer to the nearest hundredth. m/s2

1. 294 N

The weight of an object is given by:

where

m is the mass of the object

g = 9.8 m/s^2 is the acceleration of gravity

For the sled in the problem,

m = 30.0 kg

So its weight is

2. 285 N

In order to find the normal force on the sled, we need to analyze the forces acting along the vertical direction.

We have:

- Weight of the sled: W=294 N, downward

- Normal reaction: N=?, upward

- Vertical component of the pull of the child, which is given by

(upward)

So the equation of the forces is

And substituting the values we can find the magnitude of the normal force:

3.  

In order to find the acceleration of the sled, we need to analyze the forces acting along the horizontal direction. Here I assume that the two children are pulling and pushing the sled both forward, since it is not specified. We have:

- Horizontal component of the child pulling the sled:

(forward)

- Force of the child pushing the sled:

(forward)

- Force of friction:

(backward)

So the net force along the horizontal direction is

And given Newton's second law:

F = ma

where F is the net force, m the mass and a the acceleration of the sled, we can now find the acceleration:

snell's law

refractve index = sin incidence / sin refraction