A block of ice at 0 degrees C, whose mass is initially 62 kg, slides along a horizontal surface, starting at a speed of 5.48 m/s
1 answer:
The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g
<u>Explanation:</u>
Given,
Temperature, T = 0°C
Initial mass, Mi = 62kg
Speed, s = 5.48m/s
Distance, x = 26.8m
Friction is present.
Mass of ice melted = ?
We know,
The amount of energy required for the melting of ice is exactly equal to the initial kinetic energy of the block of ice
and
Therefore,
KE = 930.94 Joules
Ice melting lateral heat is 334 kJ/kg = 334000 J/kg.
Therefore, the melted mass of the ice = 930.94 / 334000 = 0.00278 kg = 2.78 g.
Thus, The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g
You might be interested in
When you're using a crowbar to lift a large rock, you are working against the force called
Gravity on Earth is what gives weight to all objects, it's defined as all things that have mass or energy are gravitated towards each other. Therefore when you're using a crowbar to lift a large rock, the weight is caused by gravity.
I hope this helps you!
Gravity on Earth is what gives weight to all objects, it's defined as all things that have mass or energy are gravitated towards each other. Therefore when you're using a crowbar to lift a large rock, the weight is caused by gravity.
I hope this helps you!
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows: