Question2. Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N , the second exerts a force of 90.0 N , friction is 12.0 N , and the mass of the third child plus wagon is 23.0 kg . (a) What is the system of interest if the acceleration of the child in the wagon is to be calculated? (See the free-body diagram.) (b) Calculate the acceleration. (c) What would the acceleration be if friction were 15.0 N ?

Studdy Solution

STEP 1

1. The forces exerted by the two children are in opposite directions.
2. Friction opposes the motion of the wagon.
3. The mass of the third child plus the wagon is 23.0 kg.
4. We are to calculate the acceleration of the child in the wagon.
5. The net force acting on the system is the sum of all forces, taking direction into account.

STEP 2

1. Identify the system of interest.
2. Calculate the net force acting on the system.
3. Use Newton's second law to calculate the acceleration.
4. Recalculate the acceleration with a different friction force.

STEP 3

Identify the system of interest.
The system of interest is the third child plus the wagon. This is because we want to calculate the acceleration of this combined mass under the influence of the forces applied by the other two children and friction.

STEP 4

Calculate the net force acting on the system.
The net force $F_{\text{net}}$ is the difference between the forces exerted by the two children minus the friction force, since they are in opposite directions:
$F_{\text{net}} = F_2 - F_1 - F_{\text{friction}}$
Substitute the given values:
$F_{\text{net}} = 90.0 \, \text{N} - 75.0 \, \text{N} - 12.0 \, \text{N}$
$F_{\text{net}} = 3.0 \, \text{N}$

STEP 5

Use Newton's second law to calculate the acceleration.
Newton's second law states that $F = ma$ , where $F$ is the net force, $m$ is the mass, and $a$ is the acceleration.
Rearrange to solve for acceleration $a$ :
$a = \frac{F_{\text{net}}}{m}$
Substitute the known values:
$a = \frac{3.0 \, \text{N}}{23.0 \, \text{kg}}$
$a \approx 0.130 \, \text{m/s}^2$

STEP 6

Recalculate the acceleration with a different friction force.
If the friction force is 15.0 N, the net force becomes:
$F_{\text{net}} = 90.0 \, \text{N} - 75.0 \, \text{N} - 15.0 \, \text{N}$
$F_{\text{net}} = 0.0 \, \text{N}$
Since the net force is zero, the acceleration is:
$a = \frac{0.0 \, \text{N}}{23.0 \, \text{kg}}$
$a = 0.0 \, \text{m/s}^2$

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