A block moving up a frictionless incline

[physicx_1]

Homework Statement

A block of mass 12 kg is projected up a frictionless incline with an initial velocity of 8m/s. It just reaches a point B and then slides down again.

A) what is the potential energy of the block at B if its potential energy at A is taken as zero?

b) what is the vertical height h of B above A?

c)what is the total amount of work done on the block by the force of gravity during the round trip from A to B and back again to A?

Homework Equations

The Attempt at a Solution

I 've thought of a few ideas:

its equal to the initial velocity
converted into enery
cant remember the formula that makes velocity into kinetic energy
but EK = EP

but I don't really know how to do this question at all.

I think its about the equations of motion?

thanks for any help

 

[Mandeep Deka]

Don't jumble up your ideas...

First see the problem from the energy point of view: When you project a particle with a given velocity up an frictionless inclined plane, you provide some kinetic energy to it (=1/2mv²). This K.E will be continuously converted into potential energy (=mgh) as the body moves upwards, and at some point it will stop when whole of its K.E has been converted to P.E. So you can definitely determine the height to which it will move and answer the following questions thereafter.

Hope you understood how to get through the question!

 

[physicx_1]

Don't jumble up your ideas...

First see the problem from the energy point of view: When you project a particle with a given velocity up an frictionless inclined plane, you provide some kinetic energy to it (=1/2mv²). This K.E will be continuously converted into potential energy (=mgh) as the body moves upwards, and at some point it will stop when whole of its K.E has been converted to P.E. So you can definitely determine the height to which it will move and answer the following questions thereafter.

Hope you understood how to get through the question!

thanks for your reply!

ok, so according to your response,

using (=1/2mv²), I get, 0.5 x 12 x 6sq. = 384 joules

is that how the first question is meant to be worked out?

if so, then you substitute 384 into mgh? and that's how you get the 2nd question?

I did that and using MGH for question 2 I have: 384joules = mgh. m=12 g=9.81 h=the unknown

so 384/12/9.81=3.26.

is that how its supposed to be?

 

[physicx_1]

can anyone help and do this 3rd question?

 

[rl.bhat]

can anyone help and do this 3rd question?

In part c, the particle returns back to A. So the net displacement is zero. Hence the work done should be...?