c: thats a lot of work!
1) suppose you are loading a 10 kg package onto a flatbed truck. should we lift it straight upward? or use the frictionless ramp? (which will be easier?) explain.
work: in physics, work (w) is done on objects by forces when the force (f) is exerted on the object over an interval where the object has a displacement (l). we call the angle between the force and the displacement vectors.
in equation form: (notice that w has to do with the average force acting over the interval)
(image of a 10kg package, a truck wheel, a 1m vertical distance, and a ramp at 30°)
2) we use the units of joules (j) for work. using the equation above, write an equivalent unit for joules.
3) lets use the concept of work to think about lifting the 10 - kg package to the top of the truck. what would be the minimum force required to lift the package straight up the distance of 1 m? using that much force, how much work would be done by the person’s force? show your work.
4) what would be the minimum force required to push the package up the frictionless incline if you apply the force parallel to the incline? what distance will you have to push the box for? how much work would the person pushing the box up the incline have to do? show work.
5) compare the force required, the distance the package moved, and the work done on the package for each scenario.
6) suppose a girl pulls a 2 - kg box that starts from rest with a horizontal force of 10 n and displaces a box 10 meters across a horizontal frictionless surface.
a) how much work is done by the girl on the box?
b) what is the acceleration of the box?
c) what is the final speed of the box at the end of those 10 meters?
7) suppose she pulls on the 2 - kg box with the same force, starting from rest again, but her force is pulling at an angle of 60° to the horizontal and displaces the box 10 m.
a) how much work is done by the girl on the box?
b) what is the acceleration of the box?
c) what is the final speed of the box at the end of those 10 meters?

Question

c: thats a lot of work!
1) suppose you are loading a 10 kg package onto a flatbed truck. should we lift it straight upward? or use the frictionless ramp? (which will be easier?) explain.
work: in physics, work (w) is done on objects by forces when the force (f) is exerted on the object over an interval where the object has a displacement (l). we call the angle between the force and the displacement vectors.
in equation form: (notice that w has to do with the average force acting over the interval)
(image of a 10kg package, a truck wheel, a 1m vertical distance, and a ramp at 30°)
2) we use the units of joules (j) for work. using the equation above, write an equivalent unit for joules.
3) lets use the concept of work to think about lifting the 10 - kg package to the top of the truck. what would be the minimum force required to lift the package straight up the distance of 1 m? using that much force, how much work would be done by the person’s force? show your work.
4) what would be the minimum force required to push the package up the frictionless incline if you apply the force parallel to the incline? what distance will you have to push the box for? how much work would the person pushing the box up the incline have to do? show work.
5) compare the force required, the distance the package moved, and the work done on the package for each scenario.
6) suppose a girl pulls a 2 - kg box that starts from rest with a horizontal force of 10 n and displaces a box 10 meters across a horizontal frictionless surface.
    a) how much work is done by the girl on the box?
    b) what is the acceleration of the box?
    c) what is the final speed of the box at the end of those 10 meters?
7) suppose she pulls on the 2 - kg box with the same force, starting from rest again, but her force is pulling at an angle of 60° to the horizontal and displaces the box 10 m.
    a) how much work is done by the girl on the box?
    b) what is the acceleration of the box?
    c) what is the final speed of the box at the end of those 10 meters?

Accepted Answer

### Question 1
# Brief Explanations:
To determine which is easier (lifting straight up or using a frictionless ramp), we analyze the force and work involved. Lifting straight up requires a force equal to the weight of the package ($F = mg$). Using a ramp, the force is reduced because the force is applied over a longer distance (the length of the ramp), but the work done (force × distance) should be the same in both cases (since work is related to the change in potential energy, $mgh$, and the ramp is frictionless, so no energy is lost to friction). The key is that the force required to push up the ramp is less than the weight of the package, so pushing up the ramp is easier.
# Answer:
Using the frictionless ramp is easier. Lifting straight up requires a force equal to the package’s weight ($F = mg$), while pushing up the frictionless ramp requires a smaller force (since the force is applied over a longer distance, and work done $W = Fd = mgh$ is the same in both cases, so a smaller force is needed for the longer distance of the ramp).

### Question 2
# Explanation:
## Step1: Recall the work formula
The work formula is $W = F \cdot d \cdot \cos	heta$, where $F$ is force, $d$ is distance, and $	heta$ is the angle between them. In SI units, force is measured in Newtons (N) and distance in meters (m).
## Step2: Determine the unit of Joule
Since $W$ (in Joules) $= F$ (in Newtons) $	imes d$ (in meters) $	imes \cos	heta$ (dimensionless), the unit of Joule (J) is equivalent to Newton - meter (N·m).
# Answer:
The equivalent unit for Joules is $	ext{Newton - meter (N·m)}$ (or $	ext{kg·m}^2/	ext{s}^2$ since $1\ N = 1\ kg·m/	ext{s}^2$, so $1\ J=1\ kg·m^2/	ext{s}^2$).

### Question 3
# Explanation:
## Step1: Find the minimum force to lift straight up
To lift the package straight up at a constant velocity (minimum force, since acceleration is zero), the force $F$ must balance the weight of the package. The weight $F_g=mg$, where $m = 10\ kg$ and $g = 9.8\ m/s^2$. So $F=mg=10\ kg	imes9.8\ m/s^2 = 98\ N$.
## Step2: Calculate the work done
The distance $d = 1\ m$, and the force is applied in the direction of displacement ($	heta = 0^{\circ}$, so $\cos	heta=1$). Using the work formula $W = Fd\cos	heta$, we substitute $F = 98\ N$, $d = 1\ m$, and $\cos(0^{\circ}) = 1$. So $W=98\ N	imes1\ m	imes1 = 98\ J$.
# Answer:
The minimum force required to lift the package straight up is $98\ N$. The work done by the person’s force is $98\ J$.

### Question 4
# Explanation:
## Step1: Find the length of the ramp
The height of the truck is $h = 1\ m$, and the ramp makes an angle of $30^{\circ}$ with the horizontal. Using the sine function, $\sin	heta=\frac{h}{L}$, where $L$ is the length of the ramp. So $L=\frac{h}{\sin	heta}=\frac{1\ m}{\sin(30^{\circ})}=\frac{1\ m}{0.5}=2\ m$.
## Step2: Find the minimum force to push up the ramp
For a frictionless incline, the minimum force $F$ applied parallel to the incline to move the package at a constant velocity (so net force is zero) is equal to the component of the weight along the incline. The component of weight along the incline is $F = mg\sin	heta$. Substituting $m = 10\ kg$, $g = 9.8\ m/s^2$, and $	heta = 30^{\circ}$, we get $F=10\ kg	imes9.8\ m/s^2	imes\sin(30^{\circ})=10	imes9.8	imes0.5 = 49\ N$.
## Step3: Calculate the work done
The distance pushed is the length of the ramp $d = L = 2\ m$. Using the work formula $W = Fd\cos	heta$, and since the force is applied parallel to the displacement ($	heta = 0^{\circ}$, $\cos	heta = 1$), $W=49\ N	imes2\ m=98\ J$.
# Answer:
- Minimum force to push up the incline: $49\ N$
- Distance to push the box: $2\ m$
- Work done: $98\ J$

### Question 5
# Brief Explanations:
- **Force Required**: Lifting straight up requires a force equal to the weight of the package ($F_{lift}=mg = 98\ N$), while pushing up the ramp requires a smaller force ($F_{ramp}=mg\sin	heta = 49\ N$).
- **Distance Moved**: Lifting straight up, the distance is the height of the truck ($d_{lift}=1\ m$). Pushing up the ramp, the distance is the length of the ramp ($d_{ramp}=2\ m$) (since $\sin	heta=\frac{h}{L}\implies L=\frac{h}{\sin	heta}$, with $h = 1\ m$ and $	heta = 30^{\circ}$, $L = 2\ m$).
- **Work Done**: In both cases, the work done is equal to the change in potential energy of the package, $W = mgh$. For the lift, $W_{lift}=F_{lift}	imes d_{lift}=98\ N	imes1\ m = 98\ J$. For the ramp, $W_{ramp}=F_{ramp}	imes d_{ramp}=49\ N	imes2\ m = 98\ J$. So the work done is the same in both scenarios, the force is smaller for the ramp but the distance is longer, and the force is larger for the lift but the distance is shorter.
# Answer:
- **Force**: Lifting straight up requires a larger force ($98\ N$) than pushing up the ramp ($49\ N$).
- **Distance**: Lifting straight up has a shorter distance ($1\ m$) than pushing up the ramp ($2\ m$).
- **Work Done**: The work done is the same ($98\ J$) in both scenarios (since work is related to the change in potential energy, $mgh$, and the ramp is frictionless, so no energy is lost to friction).

### Question 6
#### Part (a)
# Explanation:
## Step1: Recall the work formula
The work formula is $W = F\cdot d\cdot\cos	heta$. Here, the force $F = 10\ N$, distance $d = 10\ m$, and the angle between the force and displacement $	heta = 0^{\circ}$ (since the force is horizontal and the displacement is horizontal), so $\cos(0^{\circ})=1$.
## Step2: Calculate the work
Substitute the values into the formula: $W=10\ N	imes10\ m	imes1 = 100\ J$.
# Answer:
The work done by the girl on the box is $100\ J$.

#### Part (b)
# Explanation:
## Step1: Use Newton's second law
Newton's second law is $F = ma$, where $F$ is the net force, $m$ is mass, and $a$ is acceleration. The net force in the horizontal direction is the applied force $F = 10\ N$ (since the surface is frictionless, there is no frictional force). The mass $m = 2\ kg$.
## Step2: Solve for acceleration
From $F = ma$, we get $a=\frac{F}{m}=\frac{10\ N}{2\ kg}=5\ m/s^2$.
# Answer:
The acceleration of the box is $5\ m/s^2$.

#### Part (c)
# Explanation:
## Step1: Use the kinematic equation
We can use the kinematic equation $v_f^2=v_i^2 + 2ad$. The initial velocity $v_i = 0\ m/s$ (starts from rest), acceleration $a = 5\ m/s^2$, and distance $d = 10\ m$.
## Step2: Solve for final velocity
Substitute the values: $v_f^2=0+(2	imes5\ m/s^2	imes10\ m)=100\ m^2/s^2$. Take the square root: $v_f=\sqrt{100\ m^2/s^2}=10\ m/s$.
# Answer:
The final speed of the box is $10\ m/s$.

### Question 7
#### Part (a)
# Explanation:
## Step1: Recall the work formula
The work formula is $W = F\cdot d\cdot\cos	heta$. Here, $F = 10\ N$, $d = 10\ m$, and $	heta = 60^{\circ}$ (the angle between the force and displacement). $\cos(60^{\circ})=0.5$.
## Step2: Calculate the work
Substitute the values: $W = 10\ N	imes10\ m	imes0.5=50\ J$.
# Answer:
The work done by the girl on the box is $50\ J$.

#### Part (b)
# Explanation:
## Step1: Find the horizontal component of the force
The net force in the horizontal direction is $F_{net}=F\cos	heta$, since the vertical component of the force does not contribute to the horizontal motion (and the surface is frictionless, so no other horizontal forces). $F = 10\ N$, $	heta = 60^{\circ}$, so $F_{net}=10\ N	imes\cos(60^{\circ})=10\ N	imes0.5 = 5\ N$.
## Step2: Use Newton's second law
From $F_{net}=ma$, with $m = 2\ kg$, we get $a=\frac{F_{net}}{m}=\frac{5\ N}{2\ kg}=2.5\ m/s^2$.
# Answer:
The acceleration of the box is $2.5\ m/s^2$.

#### Part (c)
# Explanation:
## Step1: Use the kinematic equation
We use the kinematic equation $v_f^2=v_i^2 + 2ad$. $v_i = 0\ m/s$, $a = 2.5\ m/s^2$, $d = 10\ m$.
## Step2: Solve for final velocity
Substitute the values: $v_f^2=0+(2	imes2.5\ m/s^2	imes10\ m)=50\ m^2/s^2$. Take the square root: $v_f=\sqrt{50}\ m/s\approx7.07\ m/s$.
# Answer:
The final speed of the box is approximately $7.07\ m/s$ (or $\sqrt{50}\ m/s$).

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QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Recall the work formula

Step2: Determine the unit of Joule

Step1: Find the minimum force to lift straight up

Step2: Calculate the work done

Answer:

Question 2