A couple physics problems

  • #1
Punchlinegirl
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1. Mass M is sliding down the inclined part of a slide at a speed of 1.93 m/s. The mass stops a distance of S2= 2.1 m along the level part of the slide. The distance S1= 1.22 m and the angle theta= 34.70 deg. Calculate the coefficient of kinetic friction for the mass on the surface.
I used (1/2)mv^2 + mg(S1 sin theta)= FS2
(1/2)(M)(1.93^2) + M (9.8)(.694)= 2.1F
1.86M + 6.81M = 2.1 MuK M(9.8)
Canceling the M's and solving for mk gave me .421, which isn't right.


2.The potential energy stored in the compressed spring of a dart gun, with a spring constant of 28.00 N/m, is .880 J. Find by how much the spring is compressed. I got this part= .251 m. A .050 kg dart is fired straight up. Find the vertical distance the dart travels from its position when the spring is compressed to its highest position. I got this part= 1.794 m. The same dart is now fired horizontally from a height of 1.70 m. The dart remains in contact until the spring reaches its equilibrium position. Find the horizontal velocity of the dart at that time.
I used mgh + (1/2)kx^2= (1/2)mv^2
(.050)(9.8)(1.70) + (1/2)(28)(.251^2)= (1/2)(.050)v^2
which gave me 8.28 m/s, which isn't right

Thanks in advance for any help.
 

Answers and Replies

  • #2
Doc Al
Mentor
45,433
1,883
Punchlinegirl said:
1. Mass M is sliding down the inclined part of a slide at a speed of 1.93 m/s. The mass stops a distance of S2= 2.1 m along the level part of the slide. The distance S1= 1.22 m and the angle theta= 34.70 deg. Calculate the coefficient of kinetic friction for the mass on the surface.
I used (1/2)mv^2 + mg(S1 sin theta)= FS2
(1/2)(M)(1.93^2) + M (9.8)(.694)= 2.1F
1.86M + 6.81M = 2.1 MuK M(9.8)
Canceling the M's and solving for mk gave me .421, which isn't right.
Is the slide frictionless?
2.The potential energy stored in the compressed spring of a dart gun, with a spring constant of 28.00 N/m, is .880 J. Find by how much the spring is compressed. I got this part= .251 m. A .050 kg dart is fired straight up. Find the vertical distance the dart travels from its position when the spring is compressed to its highest position. I got this part= 1.794 m. The same dart is now fired horizontally from a height of 1.70 m. The dart remains in contact until the spring reaches its equilibrium position. Find the horizontal velocity of the dart at that time.
I used mgh + (1/2)kx^2= (1/2)mv^2
(.050)(9.8)(1.70) + (1/2)(28)(.251^2)= (1/2)(.050)v^2
which gave me 8.28 m/s, which isn't right
The dart is fired horizontally, so the spring PE transforms into KE. (The horizontal velocity does not depend on the height.)
 
  • #3
Punchlinegirl
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1. I don't think it is frictionless.. since they want the coefficent of kinetic friction.. but I'm not sure.

2. So should I use mgh + (1/2)kx^2 =2 (1/2)mv^2?
or does the spring equation just go away?
I'm kinda confused.
 
  • #4
Doc Al
Mentor
45,433
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Punchlinegirl said:
1. I don't think it is frictionless.. since they want the coefficent of kinetic friction.. but I'm not sure.
There are two sections: the incline part (S1) and the level part (S2). Your analysis assumed that the incline part was frictionless (since you did not include the work done against friction in your energy balance).
2. So should I use mgh + (1/2)kx^2 =2 (1/2)mv^2?
or does the spring equation just go away?
Picture it: You're shooting a dart out of a dart gun horizontally. To find the speed that it shoots out, use energy conservation: Initial Energy (spring PE) = Final Energy (kinetic). (Gravitational PE has nothing to do with it.)
 

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