Question

Please answer and provide an explanation with the necessary formula. Thank you in advance,

1. A 200kg block slides down a rough inclined plane of length 2.5m inclined at 20 degrees. At the bottom of the plane the speed of the block is 3.15 m/s. Initially the block was at rest. What is the change in the internal energy of the system? What is the frictional force?

2. A sled at rest in snow seated with two children with total mass of 50kg. You pull the sled with a rope that makes an angle of 42 degrees with the horizontal. The coefficient of kinetic friction is 0.150 and the tension in the rope is 150 N. What is the speed of the sled after it has moved through 3.5m?

3. Oxygen gas has a molar mass of 32 g/mol and carbon dioxide has a molar mass of 44.05 g/mol. What is the rms speed of the oxygen molecule at 300? If the carbon dioxide molecule has the same rms speed as oxygen molecule, what is the temperature of the carbon dioxide molecule?

4. A U tube contains a liquid of unknown density. An oil of density 800 kg m/s is poured into the tube in one of the arm and the oil column is 12 cm high. The air interface is 5 cm above the liquid level in the other. What is the density of the liquid?

5. A uniform ladder of weigh 75 N leans against a frictionless vertical wall. The foot of the ladder is 2.5m away from the wall. The length of the ladder is 5m. A fireman of weigh 700 N is on the ladder at a distance of 4m along the ladder. What is the magnitude and direction of the force exerted by the floor and ladder?

Answer 1

1) Here, we use the law of conservation of energy. Initially, when the block is at the top, the total energy of the system = mgh = 200*9.8*2.5*sin20 = 1675.8987J

When the block reaches down, the kinetic energy of the block is= 0.5*m*v^2 = 0.5*200*(3.15)^2 = 992.25J

So, the change in internal energy is given by the difference in the mechanical energies of the system initially and finally = 1675.8987-992.25 = 683.1487J

This change change in internal energy is the work done by friction = F*2.5 = 683.1487, So the force of friction F = 273.25948N

2) The normal force acting on the sled = 50*9.8-150sin42 = 389.630N, so the frictional force acting on the sled = 0.15*389.630 = 58.4445N.

The net force pulling the sled = 150cos42 - 58.4445 = 53.0272N, so the acceleration of the sled = 53.0272/50 = 1.0605m/s^2.

Using a formula from kinematics, v^2 - 0 = 2*1.0605*3.5, then v = 2.7246m/s

3) RMS velocity of a gas molecule is given by

for Oxygen at T=300, M = 32, we get RMS velocity of oxygen molecules = 15.29154 m/s

If CO2 molecules have the same RMS velocity, then the temperature would be

4) Here, the weight of oil in one arm should balance the weight of the unknown liquid in the other arm, So, 800*0.12 = 0.07*density of liquid.

then the density of the liquid = 9600/7 = 1371.428 Kg/m^3

5) The direction of force is horizontal towards the vertical wall.

The magnitude is calculated by equating the moments about the point of contact of the ladder with the vertical wall. Let F be the frictional force.

F*5cos30 = 700cos60*1+75cos60*2.5, we get F = 102.4827N