# Determine expressions for the following in terms of M, X, D, h and g

• paulimerci
In summary: D}{h}$$substituting ##v_{ix} = D^2 \frac{g}{2h}## in the above equation$$k = m\frac{D^2}{x^2} \frac{g}{2h}$$In the above equation x is...$$x = \frac{D}{h}$$#### paulimerci Homework Statement Attached image below! Relevant Equations Conservation of energy, kinematic equations d) The block of mass m is pushed toward the wall until the spring has been compressed, and so the spring has stored PE, and when the block is released, it follows a projectile motion and strikes the floor ##stored PE\rightarrow K.E ##. By law of conservation of energy,$$ E_i = E_f  mgh + \frac{1}{2}kx^2 = \frac {1}{2}mv_f^2 k = \frac{2m}{x^2} (\frac {v^2}{2} - gh)$$a) The time elapsed from the instant the block leaves the table to the instant it strikes the floor is taken as ##\Delta t##.$$ E_i = E_f  mgh + \frac{1}{2}kx^2 = \frac {1}{2}mv_f^2$$I replace ##v_f## in the above equation with range D over elapsed time.$$mgh + \frac{1}{2}kx^2 = \frac{1}{2}m \frac{D^2}{\Delta t^2}$$In the above equation I substitute for k which I got from part d.$$\Delta t = \sqrt {\frac {D}{V}}$$b) When the block takes a projectile motion, it has uniform motion at constant velocity in the horizontal direction. There fore ## (v_x)_f = (v_x)_i## = constant c) The work done on the block by the spring is given by$$ W.D_{sp} = \frac{1}{2}kx^2$$Substituting k in the above equation we get,$$ W.D_{sp} = m (\frac {v^2}{2} - gh)$$Have I done it right? #### Attachments • Screen Shot 2023-01-11 at 10.54.26 AM.png 103 KB · Views: 35 Doesn't look right to me. For part d first figure out the velocity of the block at the top of the table, then just apply kinematic relationships. paulimerci said: I replace vf in the above equation with range D over elapsed time. That would be right if ##v_f## is just the final horizontal velocity, but in your energy equation you used it as the entire final velocity. haruspex said: That would be right if ##v_f## is just the final horizontal velocity, but in your energy equation you used it as the entire final velocit erobz said: Doesn't look right to me. For part d first figure out the velocity of the block at the top of the table, then just apply kinematic relationships. Distance fallen by a mass is given by$$\Delta y = v_{iy} t - \frac{1}{2}gt^2\Delta y = -\frac{1}{2}gt^2$$since v_{iy} = 0 The block when released has initial constant velocity, it covers a distance D,$$ \Delta x = v_{ix}t + \frac{1}{2}at^2 \Delta x = v_{ix} t$$where a =0$$ v_{ix} = \frac{\Delta x}{t}$$paulimerci said: Distance fallen by a mass is given by$$\Delta y = v_{iy} t - \frac{1}{2}gt^2\Delta y = -\frac{1}{2}gt^2$$since v_{iy} = 0 The block when released has initial constant velocity, it covers a distance D,$$ \Delta x = v_{ix}t + \frac{1}{2}at^2 \Delta x = v_{ix} t$$where a =0$$ v_{ix} = \frac{\Delta x}{t}$$Better. Now substitute the given variables M, D, x, as appropriate and find ##v_{ix}##. haruspex said: Better. Now substitute the given variables M, D, x, as appropriate and find ##v_{ix}##. Any hints? paulimerci said: Any hints? using law of conservation of energy,$$mgh +\frac{1}{2}kx^2 = \frac{1}{2}mv_{ix}^2 +mghv_{ix} = \sqrt {\frac {k}{m}} x$$sSubstituting for ##v_{ix} =\frac {D}{t}## in the above equation we get for k is$$k =\frac{D}{t} ^2 m x$$The above equation is D squared over t squared. paulimerci said: using law of conservation of energy,$$mgh +\frac{1}{2}kx^2 = \frac{1}{2}mv_{ix}^2 +mghv_{ix} = \sqrt {\frac {k}{m}} x$$sSubstituting for ##v_{ix} =\frac {D}{t}## in the above equation we get for k is$$k =\frac{D}{t} ^2 m x$$The above equation is D squared over t squared.$$ v_{ix}t = D$$You have an expression for ##v_{ix}##, and you have an expression for ##t## from falling a distance ##h##. erobz said:$$ v_{ix}t = D$$You have an expression for ##v_{ix}##, and you have an expression for ##t## from falling a distance ##h##. I honestly don't know how to do this. It's quite tricky. paulimerci said: I honestly don't know how to do this. It's quite tricky. Expression for ##v_{ix} =\frac{D}{t}## and for ##t = \frac {D} {v_{ix}}## paulimerci said: Expression for ##v_{ix} =\frac{D}{t}## and for ##t = \frac {D} {v_{ix}}## Is it like this? paulimerci said: Is it like this? No, you have another expression which you can use to eliminate time ##t##. Its the first part of your post#4 paulimerci erobz said: No, you have another expression which you can use to eliminate time ##t##. Its the first part of your post#4 Okay,$$ h = \frac{1}{2}gt^2 t = \sqrt {\frac{2h}{g}}$$Substituting t in equation for v_{ix} we get$$ D = v_{ix} \sqrt {\frac{2h}{g}}$$paulimerci said: Okay,$$ h = \frac{1}{2}gt^2 t = \sqrt {\frac{2h}{g}}$$Substituting t in equation for v_{ix} we get$$ D = v_{ix} \sqrt {\frac{2h}{g}}$$Right. Now combine that with an equation relating ##v_{ix}## to k. paulimerci haruspex said: Right. Now combine that with an equation relating ##v_{ix}## to k.$$\frac{1}{2}kx^2 = \frac{1}{2}mv_{ix}^2$$substituting ##v_{ix} = D^2 \frac{g}{2h}## in the above equation$$k = m\frac{D^2}{x^2} \frac{g}{2h}$$In the above equation x is squared. Last edited: paulimerci said:$$\frac{1}{2}kx^2 = \frac{1}{2}mv_{ix}^2$$substituting ##v_{ix} = D^2 \frac{g}{2h}## in the above equation$$k = m\frac{D}{x}^2 \frac{g}{2h}$$In the above equation x is squared. work done on the block by spring$$W{sp} = \frac{1}{2} kx^2 W{sp} = \frac {mD^2g}{4h}$$Last edited: erobz paulimerci said: In the above equation x is squared. You can always just edit if it didn't parse like you suspected.$$ k = m\frac{D^2}{x^2} \frac{g}{2h}$$paulimerci said: substituting ##v_{ix} = D^2 \frac{g}{2h}## in the above equation You mean ##v_{ix}^2 = D^2 \frac{g}{2h}##. paulimerci said:$$k = m\frac{D}{x}^2 \frac{g}{2h}$$In the above equation x is squared. So write$$k = m(\frac{D}{x})^2 \frac{g}{2h}$$or$$k = m\frac{D^2}{x^2 }\frac{g}{2h}$$paulimerci said: work done on the block by spring$$W.D_{sp} = \frac{1}{2} kx^2 W.D_{sp} = \frac {mD^2g}{4h}$$You've lost me. Is W supposed to be the work done? What is Dsp? Sounds like it is the same as x, but why would you multiply the work with the displacement? erobz said: You can always just edit if it didn't parse like you suspected.$$ k = m\frac{D^2}{x^2} \frac{g}{2h}$$I did the same thing as you, but latex was not working. haruspex said: You mean ##v_{ix}^2 = D^2 \frac{g}{2h}##. So write$$k = m(\frac{D}{x})^2 \frac{g}{2h}$$or$$k = m\frac{D^2}{x^2 }\frac{g}{2h}$$You've lost me. Is W supposed to be the work done? What is Dsp? Sounds like it is the same as x, but why would you multiply the work with the displacement? No, no, it's just work done with W, not W.D. I'll edit it now. paulimerci said: I did the same thing as you, but latex was not working. No you didn't. You did the following: k = m\frac{D}{x}^2 \frac{g}{2h} the proper codes are: $$k = m\frac{D^2}{x^2} \frac{g}{2h}$$ ##k = m\frac{D^2}{x^2} \frac{g}{2h}## or $$k = m \left( \frac{D}{x} \right)^2 \frac{g}{2h} ## k = m \left( \frac{D}{x} \right)^2 \frac{g}{2h}##

paulimerci said:
I did the same thing as you, but latex was not working.
You didn't. Use edit on your reply to see what the LaTeX is in erobz' version and compare it with yours.

haruspex said:
You didn't. Use edit on your reply to see what the LaTeX is in erobz' version and compare it with yours.
I did it!

paulimerci said:
No, no, it's just work done with W, not W.D. I'll edit it now.
ok

Thank you so much @erobz and @haruspex for your great help. I thought I wouldn't be able to complete it and felt the problem was tricky. The reason I started finding for k with energy conservation and not Newtonian mechanics was because I thought acceleration is not constant because spring force is a varying one, which in turn leads to the understanding that acceleration is not constant in the horizontal direction.

erobz
erobz said:
Doesn't look right to me.

For part d first figure out the velocity of the block at the top of the table, then just apply kinematic relationships.
I’m not sure whether I’m saying it right. If the spring has varying force when it is released then the block’s velocity can’t be constant that means acceleration is not constant along the horizontal direction. Right?

paulimerci said:
I’m not sure whether I’m saying it right. If the spring has varying force when it is released then the block’s velocity can’t be constant that means acceleration is not constant along the horizontal direction. Right?
I believe @erobz only meant you to apply the kinematic relationships after the block leaves the table.

erobz
haruspex said:
I believe @erobz only meant you to apply the kinematic relationships after the block leaves the table.
But we find velocity in part b for horizontal direction where we take acceleration to be zero. This applies only when the block moves with constant velocity. Right?

paulimerci said:
But we find velocity in part b for horizontal direction where we take acceleration to be zero. This applies only when the block moves with constant velocity. Right?
Is the acceleration along x direction is zero or not zero?

paulimerci said:
But we find velocity in part b for horizontal direction where we take acceleration to be zero. This applies only when the block moves with constant velocity. Right?
Why "but"?
Yes, for part b we only need to consider motion after the block leaves the table, and in that phase vertical acceleration is constant, g, and horizontal acceleration is constant, 0. So we can apply the kinematic equations for constant acceleration.

paulimerci
haruspex said:
Why "but"?
Yes, for part b we only need to consider motion after the block leaves the table, and in that phase vertical acceleration is constant, g, and horizontal acceleration is constant, 0. So we can apply the kinematic equations for constant acceleration.
Yes, while the block is in parabolic motion the horizontal velocity remains constant throughout its path. I think I understood! Thank you.