Question

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

Answer 1

1. 294 N

The weight of an object is given by:

$W=mg$

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

$W=(30.0 kg)(9.8 m/s^2)=294 N$

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

$F_{1y} = (12.0 N)sin 45^{\circ}=8.5 N$ (upward)

So the equation of the forces is

$W=N+F_{1y}$

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

$N=W-F_{1y}=294 N -8.5 N=285.5 N\sim 285 N$

3.  $0.38 m/s^2$

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:

$F_{1x} = (12.0 N)cos 45^{\circ}=8.5 N$ (forward)

- Force of the child pushing the sled:

$F_2 = 8.0 N$ (forward)

- Force of friction:

$F_f = 5.0 N$ (backward)

So the net force along the horizontal direction is

$F=F_{1x}+F_2-F_f = 8.5 N +8.0 N -5.0 N =11.5 N$

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:

$a=\frac{F}{m}=\frac{11.5 N}{30.0 kg}=0.38 m/s^2$

Answer 2

Answer: 294 N, 286 N, -0.38 m/s2

Explanation: