Asked by burnout
Net Force and Acceleration Quick Check
3 of 53 of 5 Items
Question
A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to accelerate across the ice. If drag and the friction from the ice apply a force of 5 N on the boy, what is the magnitude of the net force acting on him?(1 point)
Responses
80 N
80 N
380 N
380 N
70 N
70 N
370 N
Answers
Answered by
GPT-4o mini
To find the net force acting on the boy, we can use the following formula:
\[ \text{Net Force} = \text{Total Forces Applied} - \text{Friction/Drag Force} \]
The boy exerts a force of 300 N, his friend pulls him with a force of 75 N, and there is a drag/friction force of 5 N acting against the motion.
**Total Forces Applied:**
\[ 300 \, \text{N} + 75 \, \text{N} = 375 \, \text{N} \]
**Subtract the Drag/Friction Force:**
\[ \text{Net Force} = 375 \, \text{N} - 5 \, \text{N} = 370 \, \text{N} \]
So the magnitude of the net force acting on the boy is **370 N**.
\[ \text{Net Force} = \text{Total Forces Applied} - \text{Friction/Drag Force} \]
The boy exerts a force of 300 N, his friend pulls him with a force of 75 N, and there is a drag/friction force of 5 N acting against the motion.
**Total Forces Applied:**
\[ 300 \, \text{N} + 75 \, \text{N} = 375 \, \text{N} \]
**Subtract the Drag/Friction Force:**
\[ \text{Net Force} = 375 \, \text{N} - 5 \, \text{N} = 370 \, \text{N} \]
So the magnitude of the net force acting on the boy is **370 N**.