Question

question 8
a water balloon hits a target with +7 n of force.
how much force does the target apply on the water balloon?

does the target apply the force in the positive or negative direction? explain how you know.

question 9
a 50 kg crate is sliding across a surface while experiencing a rightward applied force of 250 n. the box is accelerating at +2.8 m/s². calculate the force of friction acting on the box.

question 10
a 12 kg box is sliding across a surface while experiencing a rightward applied force of 58 n. the box is accelerating at 0.7 m/s².
part a: calculate the normal force acting on the box.

Explanation:

QUESTION 8

Step1: Recall Newton's Third Law

Newton's Third Law states that for every action, there is an equal and opposite reaction. So, if the water balloon exerts a force of \( +7 \, \text{N} \) on the target, the target exerts a force of equal magnitude but opposite direction on the water balloon.

Step2: Determine Force Magnitude and Direction

The magnitude of the force the target applies is \( 7 \, \text{N} \). Since the action force (balloon on target) is \( +7 \, \text{N} \), the reaction force (target on balloon) is \( -7 \, \text{N} \) (opposite direction). The target applies the force in the negative direction because Newton’s Third Law dictates that forces between interacting objects are equal in magnitude and opposite in direction.

Step1: Recall Newton's Second Law

Newton's Second Law is \( F_{\text{net}} = ma \), where \( F_{\text{net}} \) is the net force, \( m \) is mass, and \( a \) is acceleration. The net force is also the sum of the applied force (\( F_{\text{applied}} \)) and the frictional force (\( F_{\text{friction}} \)): \( F_{\text{net}} = F_{\text{applied}} + F_{\text{friction}} \).

Step2: Calculate Net Force

Given \( m = 50 \, \text{kg} \) and \( a = +2.8 \, \text{m/s}^2 \), calculate \( F_{\text{net}} \):
\( F_{\text{net}} = ma = 50 \times 2.8 = 140 \, \text{N} \).

Step3: Solve for Frictional Force

We know \( F_{\text{applied}} = 250 \, \text{N} \) and \( F_{\text{net}} = F_{\text{applied}} + F_{\text{friction}} \). Rearranging for \( F_{\text{friction}} \):
\( F_{\text{friction}} = F_{\text{net}} - F_{\text{applied}} = 140 - 250 = -110 \, \text{N} \). The negative sign indicates the frictional force is in the opposite direction of the applied force (leftward), and its magnitude is \( 110 \, \text{N} \).

Step1: Analyze Forces in Vertical Direction

In the vertical direction, the box is in equilibrium (no acceleration vertically), so the normal force (\( F_{\text{N}} \)) balances the weight (\( F_{\text{g}} \)) of the box. The weight of an object is given by \( F_{\text{g}} = mg \), where \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).

Step2: Calculate Weight (and Normal Force)

Given \( m = 12 \, \text{kg} \), calculate \( F_{\text{g}} \):
\( F_{\text{g}} = mg = 12 \times 9.8 = 117.6 \, \text{N} \). Since the box is in vertical equilibrium, \( F_{\text{N}} = F_{\text{g}} = 117.6 \, \text{N} \) (the normal force acts upward to balance the downward - acting weight).

Answer:

The target applies a force of \( 7 \, \text{N} \) in the negative direction. This is because of Newton’s Third Law (action - reaction), where the force exerted by the target on the balloon is equal in magnitude (\( 7 \, \text{N} \)) and opposite in direction to the force the balloon exerts on the target (\( +7 \, \text{N} \)), so it is \( -7 \, \text{N} \) (negative direction).

QUESTION 9