(1) A 100kg block is started with a speed of 2.0m/s on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20. Consider the situation from a frame of reference moving at 2.0m/s along the initial velocity of block. Find the work done in this frame by the external force holding the belt.

(2)A ball is dropped on a floor from a height of 2.0m. After the collision it rises up 2 a height of 1.5m Assume that 40% of mechanical energy lost goes as thermal energy into the ball. Calculate the rise in temperature of the ball in the collision. Heat capacity of the ball is 800J/K

(3) A metal block of density 6000kg/m^3 and mass 1.2 kg is suspended through a spring of spring constant 200N/m. The spring-block system is dipped in water kept in a vessel. The water has a mass of 260g and the block is at a height of 40cm above the bottom of the vessel. If the support 2 the spring is broken, what will be the rise in temperature of the water. Specific heat capacity of the block is 250J/kg-K and that of water is 4200J/kg-K. Heat capacities of the vessel and the spring are nelgligible.

Also if possible, please answer this questions also

http://www.goiit.com/posts/list/heat-thermodynamics-glass-tube-questions-24600.htm

Let's break down each of your questions step by step, ensuring we understand the underlying physics concepts involved. We'll tackle them one by one, starting with the first scenario involving the block on the rough belt.

1. Work Done by External Force on a Block

In this scenario, we have a 100 kg block moving at an initial speed of 2.0 m/s on a rough belt with a coefficient of kinetic friction of 0.20. To find the work done by the external force holding the belt from the frame of reference moving at the same speed as the block, we need to consider the forces acting on the block.

Calculating the Frictional Force

The frictional force (f) acting on the block can be calculated using the formula:

• f = μ * N

Where:

• μ = coefficient of kinetic friction = 0.20
• N = normal force = weight of the block = mass * gravity = 100 kg * 9.81 m/s² = 981 N

Now, substituting the values:

• f = 0.20 * 981 N = 196.2 N

Work Done by the External Force

In the frame of reference moving with the block, the block does not experience any acceleration, and thus the net force acting on it is zero. The work done by the external force holding the belt must equal the work done against friction to keep the block moving at constant speed. The work done (W) can be calculated as:

• W = f * d

However, since the block is moving at a constant speed, we can consider the distance (d) over which the friction acts. If we assume the block moves a distance of d, the work done by the external force is equal to the work done against friction:

• W = 196.2 N * d

Since we don't have a specific distance, we can express the work done in terms of distance d. Thus, the work done by the external force is:

• W = 196.2d

2. Temperature Rise of the Ball After Collision

Next, we analyze the situation with the ball dropped from a height of 2.0 m. After hitting the floor, it rises to a height of 1.5 m, losing some mechanical energy in the process.

Calculating Energy Loss

The potential energy (PE) lost when the ball falls can be calculated using:

• PE_initial = m * g * h_initial
• PE_final = m * g * h_final

Where:

• m = mass of the ball (let's assume 1 kg for simplicity)
• g = 9.81 m/s²
• h_initial = 2.0 m
• h_final = 1.5 m

Calculating the potential energies:

• PE_initial = 1 kg * 9.81 m/s² * 2.0 m = 19.62 J
• PE_final = 1 kg * 9.81 m/s² * 1.5 m = 14.715 J

The energy lost during the collision is:

• Energy_lost = PE_initial - PE_final = 19.62 J - 14.715 J = 4.905 J

Calculating Temperature Rise

Given that 40% of this energy loss goes into heating the ball:

• Energy_to_ball = 0.40 * 4.905 J = 1.962 J

The rise in temperature (ΔT) can be calculated using the formula:

• ΔT = Energy_to_ball / Heat_capacity

Substituting the values:

• ΔT = 1.962 J / 800 J/K = 0.0024525 K

Thus, the rise in temperature of the ball is approximately 0.00245 K.

3. Temperature Rise of Water After Spring Break

Now, let’s consider the metal block suspended through a spring that is dipped in water. When the spring support is broken, the block will fall and transfer its energy to the water.

Calculating Energy of the Block

The potential energy of the block when it is at a height of 40 cm can be calculated as:

• PE_block = m * g * h = 1.2 kg * 9.81 m/s² * 0.4 m = 4.7052 J

Heat Transfer to Water

When the block falls, it will lose this potential energy, which will be transferred to the water. Assuming all the energy goes into heating the water, we can calculate the rise in temperature of the water (ΔT_water) using:

• ΔT_water = Energy_block / (mass_water * specific_heat_water)

Given:

• mass_water = 0.260 kg
• specific_heat_water = 4200 J/kg-K

Substituting the values:

• ΔT_water = 4.7052 J / (0.260 kg * 4200 J/kg-K) = 4.7052 / 1092 = 0.0043 K

Thus, the rise in temperature of the water is approximately 0.0043 K.

Additional Questions

For the additional questions regarding heat and thermodynamics in glass tubes, I recommend checking the specific details provided in the link you mentioned. If you have specific scenarios or problems from that resource, feel free to share them, and I can help you work through those as