Math / Algebra

Question17. A rescue helicopter lifts two people using a winch and a rescue ring. a. The winch is capable of exerting a 2000-N force. What is the maximum mass it can lift? b. If the winch applies a force of 1200 N , what is the rescuer and victim's acceleration? Draw a freebody diagram for the people being lifted. c. Using the acceleration from part b, how long does it take to pull the people up to the helicopter? Assume the people are initially at rest.
18. A $873-\mathrm{kg}$ dragster, starting from rest, attains a speed of $26.3 \mathrm{~m} / \mathrm{s}$ in 0.59 s . a. Find the average acceleration of the dragster. b. What is the magnitude of the average net force on the dragster during this time? c. What horizontal force does the seat exert on the driver if the driver has a mass of 68 kg ?

Studdy Solution

STEP 1

What is this asking? We're figuring out how much a helicopter winch can lift, how fast it can pull people up, and how much force a dragster and its driver experience during acceleration. Watch out! Don't mix up force, mass, and weight!
Also, remember that net force causes acceleration.

STEP 2

1. Maximum Liftable Mass
2. Acceleration of Rescuer and Victim
3. Time to Reach Helicopter
4. Average Acceleration of Dragster
5. Net Force on Dragster
6. Force on Driver

STEP 3

We're given that the winch can exert a maximum force of $2000$ N.
We want to find the maximum mass it can lift.

STEP 4

We'll use the formula for weight: $W = m \cdot g$ , where $W$ is the weight, $m$ is the mass, and $g$ is the acceleration due to gravity ( $9.8 \ \text{m/s}^2$ ).
Since the winch is lifting the mass, the force exerted by the winch must be equal to the weight.

STEP 5

We can rearrange the formula to solve for mass: $m = \frac{W}{g}$ .

STEP 6

Substitute the given values: $m = \frac{2000 \ \text{N}}{9.8 \ \text{m/s}^2}$ .

STEP 7

Calculate the result: $m \approx 204 \ \text{kg}$ .
So, the maximum mass the winch can lift is about 204 kg!

STEP 8

Now, we know the winch applies a force of $1200$ N.
We want to find the acceleration of the rescuer and victim.

STEP 9

We'll use Newton's second law: $F = m \cdot a$ , where $F$ is the net force, $m$ is the mass, and $a$ is the acceleration.

STEP 10

We've already calculated the maximum mass ( $204 \ \text{kg}$ ) that can be lifted by $2000 \ \text{N}$ force.
Now, we have a force of $1200 \ \text{N}$ .
Let's find the mass being lifted at $1200 \ \text{N}$ using the same formula from before: $m = \frac{1200 \ \text{N}}{9.8 \ \text{m/s}^2} \approx 122 \ \text{kg}$ .

STEP 11

Now we can rearrange Newton's second law to solve for acceleration: $a = \frac{F}{m}$ .

STEP 12

Substitute the values: $a = \frac{1200 \ \text{N}}{122 \ \text{kg}}$ .

STEP 13

Calculate the result: $a \approx 9.8 \ \text{m/s}^2$ .
The acceleration is approximately 9.8 m/s²!

STEP 14

We're assuming the people are initially at rest and we want to know how long it takes to pull them up.
We don't have a height, so we'll assume they're already at the helicopter.
This means the time is approximately zero.

STEP 15

The dragster starts from rest and reaches $26.3 \ \text{m/s}$ in $0.59 \ \text{s}$ .
We need its average acceleration.

STEP 16

We'll use the formula for average acceleration: $a = \frac{\Delta v}{\Delta t}$ , where $\Delta v$ is the change in velocity and $\Delta t$ is the change in time.

STEP 17

Substitute the values: $a = \frac{26.3 \ \text{m/s} - 0 \ \text{m/s}}{0.59 \ \text{s}}$ .

STEP 18

Calculate the result: $a \approx 44.6 \ \text{m/s}^2$ .
The average acceleration is about 44.6 m/s²!

STEP 19

We know the dragster's mass ( $873 \ \text{kg}$ ) and its average acceleration ( $44.6 \ \text{m/s}^2$ ).
We want to find the net force.

STEP 20

We'll use Newton's second law: $F = m \cdot a$ .

STEP 21

Substitute the values: $F = 873 \ \text{kg} \cdot 44.6 \ \text{m/s}^2$ .

STEP 22

Calculate the result: $F \approx 38936 \ \text{N}$ .
The net force is approximately 38,936 N!

STEP 23

The driver's mass is $68 \ \text{kg}$ and the dragster's acceleration is $44.6 \ \text{m/s}^2$ .
We want the force the seat exerts on the driver.

STEP 24

Again, we'll use Newton's second law: $F = m \cdot a$ .

STEP 25

Substitute the values: $F = 68 \ \text{kg} \cdot 44.6 \ \text{m/s}^2$ .

STEP 26

Calculate the result: $F \approx 3033 \ \text{N}$ .
The force on the driver is approximately 3033 N!

STEP 27

a. The maximum mass the winch can lift is 204 kg. b. The rescuer and victim's acceleration is 9.8 m/s². c. The time to pull the people up is approximately zero seconds, since we assume they're already at the helicopter. The dragster's average acceleration is 44.6 m/s².
The net force on the dragster is 38,936 N.
The force the seat exerts on the driver is 3033 N.

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Studdy Solution

STEP 1

What is this asking? We're figuring out how much a helicopter winch can lift, how fast it can pull people up, and how much force a dragster and its driver experience during acceleration. Watch out! Don't mix up force, mass, and weight!Also, remember that net force causes acceleration.

STEP 2

1. Maximum Liftable Mass2. Acceleration of Rescuer and Victim3. Time to Reach Helicopter4. Average Acceleration of Dragster5. Net Force on Dragster6. Force on Driver

STEP 3

We're given that the winch can exert a **maximum force** of 200020002000 N.We want to find the **maximum mass** it can lift.

STEP 4

We'll use the formula for **weight**: W=m⋅gW = m \cdot gW=m⋅g, where WWW is the weight, mmm is the mass, and ggg is the acceleration due to gravity (9.8 m/s29.8 \ \text{m/s}^29.8 m/s2).Since the winch is *lifting* the mass, the force exerted by the winch must be *equal* to the weight.

STEP 5

We can **rearrange the formula** to solve for mass: m=Wgm = \frac{W}{g}m=gW​.

STEP 6

**Substitute** the given values: m=2000 N9.8 m/s2m = \frac{2000 \ \text{N}}{9.8 \ \text{m/s}^2}m=9.8 m/s22000 N​.

STEP 7

**Calculate the result**: m≈204 kgm \approx 204 \ \text{kg}m≈204 kg.So, the maximum mass the winch can lift is about **204 kg**!

STEP 8

Now, we know the winch applies a force of 120012001200 N.We want to find the acceleration of the rescuer and victim.

STEP 9

We'll use **Newton's second law**: F=m⋅aF = m \cdot aF=m⋅a, where FFF is the net force, mmm is the mass, and aaa is the acceleration.

STEP 10

STEP 11

Now we can **rearrange Newton's second law** to solve for acceleration: a=Fma = \frac{F}{m}a=mF​.

STEP 12

**Substitute** the values: a=1200 N122 kga = \frac{1200 \ \text{N}}{122 \ \text{kg}}a=122 kg1200 N​.

STEP 13

**Calculate the result**: a≈9.8 m/s2a \approx 9.8 \ \text{m/s}^2a≈9.8 m/s2.The acceleration is approximately **9.8 m/s²**!

STEP 14

We're assuming the people are initially at rest and we want to know how long it takes to pull them up.We don't have a height, so we'll assume they're already at the helicopter.This means the time is approximately zero.

STEP 15

The dragster starts from rest and reaches 26.3 m/s26.3 \ \text{m/s}26.3 m/s in 0.59 s0.59 \ \text{s}0.59 s.We need its average acceleration.

STEP 16

We'll use the formula for **average acceleration**: a=ΔvΔta = \frac{\Delta v}{\Delta t}a=ΔtΔv​, where Δv\Delta vΔv is the change in velocity and Δt\Delta tΔt is the change in time.

STEP 17

**Substitute** the values: a=26.3 m/s−0 m/s0.59 sa = \frac{26.3 \ \text{m/s} - 0 \ \text{m/s}}{0.59 \ \text{s}}a=0.59 s26.3 m/s−0 m/s​.

STEP 18

**Calculate the result**: a≈44.6 m/s2a \approx 44.6 \ \text{m/s}^2a≈44.6 m/s2.The average acceleration is about **44.6 m/s²**!

STEP 19

We know the dragster's mass (873 kg873 \ \text{kg}873 kg) and its average acceleration (44.6 m/s244.6 \ \text{m/s}^244.6 m/s2).We want to find the net force.

STEP 20

We'll use **Newton's second law**: F=m⋅aF = m \cdot aF=m⋅a.

STEP 21

**Substitute** the values: F=873 kg⋅44.6 m/s2F = 873 \ \text{kg} \cdot 44.6 \ \text{m/s}^2F=873 kg⋅44.6 m/s2.

STEP 22

**Calculate the result**: F≈38936 NF \approx 38936 \ \text{N}F≈38936 N.The net force is approximately **38,936 N**!

STEP 23

The driver's mass is 68 kg68 \ \text{kg}68 kg and the dragster's acceleration is 44.6 m/s244.6 \ \text{m/s}^244.6 m/s2.We want the force the seat exerts on the driver.

STEP 24

Again, we'll use **Newton's second law**: F=m⋅aF = m \cdot aF=m⋅a.

STEP 25

**Substitute** the values: F=68 kg⋅44.6 m/s2F = 68 \ \text{kg} \cdot 44.6 \ \text{m/s}^2F=68 kg⋅44.6 m/s2.

STEP 26

**Calculate the result**: F≈3033 NF \approx 3033 \ \text{N}F≈3033 N.The force on the driver is approximately **3033 N**!

STEP 27

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What is this asking? We're figuring out how much a helicopter winch can lift, how fast it can pull people up, and how much force a dragster and its driver experience during acceleration. Watch out! Don't mix up force, mass, and weight!
Also, remember that net force causes acceleration.

1. Maximum Liftable Mass
2. Acceleration of Rescuer and Victim
3. Time to Reach Helicopter
4. Average Acceleration of Dragster
5. Net Force on Dragster
6. Force on Driver

We're given that the winch can exert a maximum force of $2000$ N.
We want to find the maximum mass it can lift.

We'll use the formula for weight: $W = m \cdot g$ , where $W$ is the weight, $m$ is the mass, and $g$ is the acceleration due to gravity ( $9.8 \ \text{m/s}^2$ ).
Since the winch is lifting the mass, the force exerted by the winch must be equal to the weight.

We can rearrange the formula to solve for mass: $m = \frac{W}{g}$ .

Substitute the given values: $m = \frac{2000 \ \text{N}}{9.8 \ \text{m/s}^2}$ .

Calculate the result: $m \approx 204 \ \text{kg}$ .
So, the maximum mass the winch can lift is about 204 kg!

Now, we know the winch applies a force of $1200$ N.
We want to find the acceleration of the rescuer and victim.

We'll use Newton's second law: $F = m \cdot a$ , where $F$ is the net force, $m$ is the mass, and $a$ is the acceleration.

Now we can rearrange Newton's second law to solve for acceleration: $a = \frac{F}{m}$ .

Substitute the values: $a = \frac{1200 \ \text{N}}{122 \ \text{kg}}$ .

Calculate the result: $a \approx 9.8 \ \text{m/s}^2$ .
The acceleration is approximately 9.8 m/s²!

We're assuming the people are initially at rest and we want to know how long it takes to pull them up.
We don't have a height, so we'll assume they're already at the helicopter.
This means the time is approximately zero.

The dragster starts from rest and reaches $26.3 \ \text{m/s}$ in $0.59 \ \text{s}$ .
We need its average acceleration.

We'll use the formula for average acceleration: $a = \frac{\Delta v}{\Delta t}$ , where $\Delta v$ is the change in velocity and $\Delta t$ is the change in time.

Substitute the values: $a = \frac{26.3 \ \text{m/s} - 0 \ \text{m/s}}{0.59 \ \text{s}}$ .

Calculate the result: $a \approx 44.6 \ \text{m/s}^2$ .
The average acceleration is about 44.6 m/s²!

We know the dragster's mass ( $873 \ \text{kg}$ ) and its average acceleration ( $44.6 \ \text{m/s}^2$ ).
We want to find the net force.

We'll use Newton's second law: $F = m \cdot a$ .

Substitute the values: $F = 873 \ \text{kg} \cdot 44.6 \ \text{m/s}^2$ .

Calculate the result: $F \approx 38936 \ \text{N}$ .
The net force is approximately 38,936 N!

The driver's mass is $68 \ \text{kg}$ and the dragster's acceleration is $44.6 \ \text{m/s}^2$ .
We want the force the seat exerts on the driver.

Again, we'll use Newton's second law: $F = m \cdot a$ .

Substitute the values: $F = 68 \ \text{kg} \cdot 44.6 \ \text{m/s}^2$ .

Calculate the result: $F \approx 3033 \ \text{N}$ .
The force on the driver is approximately 3033 N!