Solving Physics Problems: Tension, Speed, and Resistive Force

[Raider.]

Good day, ladies and gentlemen. I am a long time board reader (since my high school days) but have not posted in these forums, till now. I initially registered with plans and hopes to enter a career in physics, now I am taking my first formal college level physics course, in the summer after my second year at university, as a student pursuing premedical studies in neurochemistry and physiology. How things have changed!

I am preparing for an exam, and in the course of 60 or so problems in the chapter on Energy, I've run into a couple pickles, problems that, while aren't particularly difficult, are of a nature where i must certainly be missing some sort of pivotal fact, relationship, or plan regarding set-up, as I've simply not been able to solve them satisfactorily. I'd appreciate any insight you bright individuals may be able to provide with these 3 problems. I don't necessarily need them worked out, just ideas as to how get moving and proceed with them. I'd be greatly in your debt, thank you.

1.) A sledge loaded with bricks has a total mass of 18.0 kg and is pulled at constant speed by a rope inclined at 20.0 degrees above the horizontal. The sledge moves a distance of 20.0 m on a horizontal surface. The coefficient of kinetic friction between the sledge and surface is .500
a.) What is the tension in the rope?
b.) How much work is done by the rope on the sledge?
c.) What is the mechanical energy lost due to friction?

I've got a good idea on how to solve parts B and C, but i simply fail to understand how I can determine the tension in the rope from the information given!

2.) Tarzan swings on a 30.0-m-long vine initially inclined at an angle of 37.0 degrees with the vertical. What is his speed at the bottom of the swing
(a) if he starts from rest?
(b) if he pushes off with a speed of 4.00 m/s?

3.)A 70.0 kg diver steps off a 10m tower and drops from rest straight down into the water. If he comes to rest 5.0m beneath the surface, determine the average resistive force exerted on him by the water.Your help is greatly appreciated, thank you!

 

[Ed Aboud]

Hi Raider!

For question 1 a) resolve the sledges weight into two vectors, one parallel and one perpendicular to the plane. Now the question states that the sledge is moving at a constant speed, therefore it is not accelerating, which means there is no net force on the sledge.
For question 2 try and use the principle of conservation of energy, i.e. the sum of tarzan's potential and kinetic energy at any moment is always the same. So take two positions, one when he starts his swing and one when he finishes his swing. For each calculate his potential and kinetic energy.
For question 3, first calculate the speed he hits the water with ( use a kinematic equation ) then try to examine the forces acting on the diver as he enters the water. First there will be his weight and the resisting force opposing his weight. You can now solve this by using Newtons second law.
Hope this helps.

 

[Moetasim]

As a fellow scientist, I am happy to assist you with these physics problems. It is great to see your enthusiasm for the subject and your determination to excel in your studies.

1.) To determine the tension in the rope, we can use the fact that the sledge is moving at a constant speed. This means that the net force acting on the sledge must be zero. Since the sledge is moving horizontally, the vertical component of the tension in the rope must be equal to the weight of the sledge (mg). Using trigonometry, we can find the horizontal component of the tension, which is also equal to the force of kinetic friction (μkN). Therefore, we can set up the following equation:

Tcos(20°) - μkmg = 0

Solving for T, we get T = μkmg/cos(20°). Plugging in the values given, we get T = (0.5)(18.0 kg)(9.8 m/s^2)/cos(20°) = 88.6 N.

2.) To find Tarzan's speed at the bottom of the swing, we can use the conservation of energy principle. At the top of the swing, Tarzan has all potential energy (mgh) and no kinetic energy. At the bottom of the swing, he has all kinetic energy (1/2mv^2) and no potential energy. Therefore, we can set up the following equation:

mgh = 1/2mv^2

Solving for v, we get v = √(2gh). Plugging in the values given, we get v = √(2)(9.8 m/s^2)(30.0 m) = 24.0 m/s.

(a) If Tarzan starts from rest, his initial potential energy is zero and his final kinetic energy is also zero. Therefore, his speed at the bottom of the swing is also zero.

(b) If Tarzan pushes off with a speed of 4.00 m/s, his initial kinetic energy is 1/2mv^2 = 1/2(70.0 kg)(4.00 m/s)^2 = 560 J. Using the same equation as above, we can solve for v and get v = √(2)(9.8 m/s^2)(30.0 m) + 560 J = 25.9 m/s