A 50.0-kg block and a 100-kg block are connected by a string...

Name: Problem 74, Chapter 8: Fundamentals of Physics
Uploaded: 2023-10-08T14:38:36-08:00
Duration: 6 min 51 s
Channel: Vishal Gupta
Description: A 50.0-kg block and a 100-kg block are connected by a string as in Figure P 8.74. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the 50.0-kg block and the incline is μk=0.250 . Determine the change in the kinetic energy of the 50.0-kg block as it moves from (A) to (B), a distance of 20.0 m .

A $50.0-\mathrm{kg}$ block and a $100-\mathrm{kg}$ block are connected by a string as in Figure $\mathrm{P} 8.74$. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the $50.0-\mathrm{kg}$ block and the incline is $\mu_{k}=0.250 .$ Determine the change in the kinetic energy of the $50.0-\mathrm{kg}$ block as it moves from (A) to (B), a distance of $20.0 \mathrm{~m} .$

Key Concepts

Work-Energy Theorem

This principle states that the net work done on an object is equal to its change in kinetic energy. In problems involving moving blocks, this theorem allows for the calculation of kinetic energy changes by summing the work done by all forces acting on the object, including gravitational forces, friction, and tension, if applicable.

Non-Conservative Forces (Friction)

Friction is a non-conservative force that converts mechanical energy into thermal energy, reducing the total mechanical energy available for kinetic or potential energy. The work done by kinetic friction, which is dependent on the coefficient of kinetic friction and the normal force, is negative and plays a crucial role in determining the overall energy change in systems involving motion on surfaces.

Gravitational Work and Potential Energy

When an object moves along an inclined plane, gravitational forces do work on it. This involves resolving the weight of the object into components parallel and perpendicular to the incline. The work done by gravity can result in a change in gravitational potential energy, which influences the net work and consequently the change in kinetic energy.

Force Resolution on an Inclined Plane

Analyzing motion on an inclined plane requires the resolution of forces into components along axes parallel and perpendicular to the surface. This technique is key to correctly identifying the contributions of gravity and friction to the work done on the object, thus enabling an accurate application of the work-energy theorem in such systems.

Transcript

00:01 Here in this given problem, this is the rough inclined plane which is inclined at an angle of 37 degree with the horizontal.

00:12 Means this angle theta, this is given to be 37 degree out of the two blocks.

00:24 The first block of mass m1 where m1 is 50 kg that is kept over the inclined plane and another block that is hanging with the help of a thread, a light string.

00:42 The string is passing over a pulley, massless frictionless pulley.

00:47 Its mass m2 where this m2 is 100 kg.

00:56 Its weight m2g acting vertically down, tension t in the string.

01:02 Weight of the first block m1g that is also acting vertically down.

01:08 Its two components, first component perpendicular to the inclined plane that is m1g cos theta and another component along the inclined plane m1g sin theta.

01:24 Normal reaction of the inclined plane over the first block n which using newton's third law of motion that should be equal to m1g cos theta.

01:39 As this block will be falling down suppose with an acceleration a, so 50 kg block that will be moving up and the two points over the inclined plane a and b.

01:56 Coefficient of kinetic friction 0 .250, the gap between the two points a and b, this is ab is equal to 20 .0 meter.

02:08 First of all using free body diagram of the hanging block for which m2g should be more, m2 that is 100 kg.

02:32 So 100g -t, this net force should be equal to mass of the block 100 kg with its multiplied by the acceleration and that is using newton's second law of motion.

02:47 This is equation number 1.

02:49 Then using free body diagram of the block kept over the inclined plane, free body diagram of 50 kg block.

03:07 As it is moving up, so tension t will be more.

03:11 Then m1g sin theta means for m1 this is 50g sin 37 degree minus force of friction acting backward fk and that should be equal to m1 into a means 50a...

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