Question

The figure shows a block slides along a frictionless track from point a to polut b. The work (W) done on the box by the gravi
0 0
Add a comment Improve this question Transcribed image text
Answer #1

In both the paths energy at the starting point (a) is same. And reached same point (b) i.e., energy is same.

In path-1, initially reached bottom most position then reached point (b) against gravity

So, net workdone = m x g x (a-b)

In path-2, intially moving in horizontal direction, gravity is same so no workdone, after it is moving vertically downwards gue to gravity.

Workdone = m x g x (a-b)

So, answer is W1 = W2

Add a comment
Know the answer?
Add Answer to:
The figure shows a block slides along a frictionless track from point a to polut b....
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • A block slides from rest, along a track with an elevated left end, a flat central...

    A block slides from rest, along a track with an elevated left end, a flat central part, into a relaxed spring, as shown in the figure. The curved portion of the track is frictionless, as well as the first portion of the flat part of L = 10 cm. The coefficient of kinetic friction between the block and the only rough part, D = 10 cm, is given by k = 0.20. Let the initial height of the block be...

  • A small block of mass m slides along the frictionless loop the loop track shown below....

    A small block of mass m slides along the frictionless loop the loop track shown below. If it starts from rest at point A, what is the speed of the block at point B? (v = squareroot (10 g R)) What is the net force acting on the block at point C? (Don't forget the gravitational force. (F = -mg (8i + j) At what height above the bottom should the block be released so that the normal force exerted...

  • In the figure, a 3.7 kg block slides along a track from one level to a...

    In the figure, a 3.7 kg block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed is v0 = 5.1 m/s, the height difference is h = 1.2 m, and μk = 0.585. Find d.

  • In the figure, a 3.6 kg block slides along a track from one level to a...

    In the figure, a 3.6 kg block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed is vo = 5.9 m/s, the height difference is h = 1.1 m, and Uk = 0.625. Find d. u=0

  • A block of mass m = 7.40 kg is released from rest from point and slides...

    A block of mass m = 7.40 kg is released from rest from point and slides on the frictionless track shown in the figure below. (Let h_a = 5.20 m.) Determine the block's speed at points Where is the energy stored at point A? Where is the energy stored at point B? m/s Determine the net work done by the gravitational force on the block as it moves from point to point

  • A block of mass m = 6.20 kg is released from rest from point and slides...

    A block of mass m = 6.20 kg is released from rest from point and slides on the frictionless track shown in the figure below. (Assume ha = 6.10 m.) Determine the block's speed at points and point m/s point m/s Determine the net work done by the gravitational force on the block as it moves from point to point J

  • A block of mass m= 7.30 kg is released from rest from point and slides on...

    A block of mass m= 7.30 kg is released from rest from point and slides on the frictionless track shown in the figure below. (Acum , -7.50m) and (a) Determine the block's speed at points point point m/s to point (b) Determine the network done by the gravitational force on the block as it moves from point

  • A block of mass m = 7.50 kg is released from rest from point and slides...

    A block of mass m = 7.50 kg is released from rest from point and slides on the frictionless track shown in the figure below. (Let ha-6.10 m.) ha 3.20 m 2.00 m (a) Determine the block's speed at points ) and vg= YC= m/s m/s (b) Determine the net work done by the gravitational force on the block as it moves from point to point C

  • 1. -/10 points SerPSE10 8.2.P.003. and slides on the frictionless track shown in the figure below....

    1. -/10 points SerPSE10 8.2.P.003. and slides on the frictionless track shown in the figure below. (Leth, = 6.70 m.) A block of mass m - 4,40 kg is released from rest from point 3.20 m 12.00 m and © (a) Determine the block's speed at points V- m/s m/s Ve= (b) Determine the network done by the gravitational force on the block as it moves from point to point ©. Need Help? Read it Watch it Submit Answer

  • A small block with mass 0.0375 kg slides in a vertical circle of radius 0.600 m...

    A small block with mass 0.0375 kg slides in a vertical circle of radius 0.600 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force exerted on the block by the track has magnitude 4.05 N . In this same revolution, when the block reaches the top of its path, point B, the magnitude of the...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT