Calculate the work done by the sliding the piano

In summary, the conversation covers two physics problems. The first problem involves calculating the work done by sliding a piano up a frictionless plank at a constant speed. The second problem involves finding the average force resisting the motion of a car driven at a constant speed with a certain power output. The conversation also includes discussions on using geometry to find the tangential component of force and the relationship between force, velocity, and mechanical power. The person asking the questions is seeking help before their upcoming test.
  • #1
SilverFalcon
1
0
Hi, this is my first post on here hope I can answer some in the future. I've finished everything except these two problems. I've got the answers except I can't figure out how to arrive at the conclusions.

1) A 4.2-kN piano is to be slid up a 3.5-m frictionless plank at a constant speed. The plank makes an angle of 30° with the horizontal. Calculate the work done by the sliding the piano up the plank. A: 74.103 J

I know I have to use this f=f(parallel)/angle and w=f*d but other than that I'm not too sure.

2) A car is driven at a constant speed of 76 km/h down a road. The car's engine delivers 48 kW of power. Calculate the average force that is resisting the motion of the car. A: 23000 N

This one seems too easy for me to be asking am I missing something really obvious?

Thanks for looking and hopfully you can help me, btw my test is tommorow, heh. :yuck:
 
Physics news on Phys.org
  • #2
SilverFalcon said:
Hi, this is my first post on here hope I can answer some in the future. I've finished everything except these two problems. I've got the answers except I can't figure out how to arrive at the conclusions.

1) A 4.2-kN piano is to be slid up a 3.5-m frictionless plank at a constant speed. The plank makes an angle of 30° with the horizontal. Calculate the work done by the sliding the piano up the plank. A: 74.103 J

I know I have to use this f=f(parallel)/angle and w=f*d but other than that I'm not too sure.

Make the drawing and then use the geometry to find the tangential component of the force...

SilverFalcon said:
2) A car is driven at a constant speed of 76 km/h down a road. The car's engine delivers 48 kW of power. Calculate the average force that is resisting the motion of the car. A: 23000 N

This one seems too easy for me to be asking am I missing something really obvious?

What is the relation between force,velocity & mechanical power...??BTW i think your answer is off a size order...

Daniel.

Thanks for looking and hopfully you can help me, btw my test is tommorow, heh. :yuck:[/QUOTE]
 
  • #3


Hi there, great job on finishing everything except for these two problems! Let's work through them together.

1) To calculate the work done by sliding the piano, we need to use the formula W = Fd, where W is the work, F is the force, and d is the distance. In this case, we have the force (4.2 kN) and the distance (3.5 m), but we need to find the actual force that is acting on the piano.

Since the plank is frictionless, the only force acting on the piano is its weight, which is equal to its mass (m) times the acceleration due to gravity (g). We can find the mass of the piano by dividing its weight (4.2 kN) by the acceleration due to gravity (9.8 m/s^2). This gives us a mass of approximately 428.6 kg.

Now, we can find the force that is acting on the piano by using the formula F = mg, where m is the mass we just calculated and g is the acceleration due to gravity. This gives us a force of approximately 4.2 kN.

Next, we need to find the component of this force that is parallel to the plank. To do this, we use the formula F(parallel) = F * cosθ, where θ is the angle between the force and the plank. In this case, θ is 30°, so we have F(parallel) = 4.2 kN * cos30° = 3.64 kN.

Finally, we can calculate the work done by sliding the piano by using the formula W = F(parallel) * d, where d is the distance the piano is being slid up the plank. In this case, d is 3.5 m. So, W = 3.64 kN * 3.5 m = 12.74 kJ. However, the question asks for the answer in joules, so we need to convert kJ to J by multiplying by 1000. This gives us a final answer of W = 12.74 kJ * 1000 = 12,740 J.

2) This problem is a bit easier since we are given the power (P) and the speed (v) of the car. We can use the formula P = Fv, where P is the power, F is the force, and
 

1. How do you calculate the work done by sliding a piano?

To calculate the work done by sliding a piano, you need to multiply the force applied to the piano by the distance it was moved. The equation for work is: Work = Force x Distance.

2. What is the unit of measurement for work?

The unit of measurement for work is joules (J). This is the same unit used to measure energy.

3. Is the work done by sliding a piano positive or negative?

The work done by sliding a piano can be either positive or negative, depending on the direction of the force and the direction of the displacement. If the force and displacement are in the same direction, the work done is positive. If they are in opposite directions, the work done is negative.

4. Can you calculate the work done by sliding a piano if the force is not constant?

Yes, you can still calculate the work done by sliding a piano if the force is not constant. In this case, you would need to use integration to find the work done. The equation for work with varying force is: Work = ∫ F(x) dx, where F(x) is the varying force and dx is the displacement.

5. How does friction affect the work done by sliding a piano?

Friction can decrease the amount of work done by sliding a piano as it opposes the motion and reduces the displacement. This means that the force applied would need to be greater to overcome the frictional force and move the piano the same distance, resulting in more work being done.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
8K
  • Introductory Physics Homework Help
Replies
14
Views
5K
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
333
Replies
1
Views
1K
  • Introductory Physics Homework Help
2
Replies
48
Views
6K
  • Introductory Physics Homework Help
Replies
2
Views
4K
Back
Top