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?

(blank) N

What is the normal force exerted on the sled? Round the answer to the nearest whole number.

(blank) N

What is the acceleration of the sled? Round the answer to the nearest hundredth.

(blank) m/s2

Answer 1

Step-by-step explanation:

Find the diagram of the scenario in the attached file.

From the diagram, the weight = mass * acceleration due to gravity

mass of sled = 30.0kg

acc. due to gravity = 9.81m/s

weight of the sled = 30.0*9.81

weight of the sled = 294.3N

The normal reaction force R in the diagram will be gotten by resolving the 12N force along the vertical

Ry = 12sin45°

R = 12 * 1/√2

R = 12√2/√2

R = 6√2 N

R = 8.49N ≈ 9.0N

The normal force exerted on the sled is approximately is 9.0N

Get the acceleration

Using the formula
`\sum Fx = ma_x` where;

m is the mass of the sled = 30.0kg

`a_x` is the acceleration of the sled

`\sum Fx = 8 + 12cos45^0\\\sum Fx = 8 + 12(0.7071)\\\sum Fx = 8 + 8.49\\\sum Fx = 16.49N`

Substitute into the formula;

`a_x = (\sum Fx )/(m) \\ a_x = (16.49)/(30)\\ a_x = 0.55 m/s^2`

Hence the acceleration of the sled rounded to the nearest hundredth is 0.55m/s²