Inclined plane and elapsed time

[Sassenav22]

Homework Statement

A block is given a initial speed of 5.0 m/s up the 20° plane. (a) How far up the plane will it go?
(b) how much much time elapses before it returns to its starting point?

Homework Equations

I think

V= u + at
s=ut t 1/2at^2
v=d/t

The Attempt at a Solution

(a) V= u + at
0= 5-9.8(sin20)t
t = 1.65s

s=ut+1/2at^2
=5(sin20)(1.65)+ 1/2(9.8)(1.65^2)
s= 16.16m

(b) v=d/t
t=d/v
=16.16/5
t= 3.2s


I know this is wrong but i can't figure it out can someone help me?

 

[Doc Al]

The Attempt at a Solution

V= u + at
0= 5-9.8(sin20)t
t = 1.65s

Recheck your arithmetic. (Tip: Once you've solved for the time the block takes to rise, what's the answer to part b?)

s=ut+1/2at^2
=5(sin20)(1.65)+ 1/2(9.8)(1.65^2)
s= 16.16m

The initial speed up the ramp is already parallel to the incline, so there's no need to multiply it by sin20. On the other hand, the acceleration along the incline is only a component of g, so you do need the sin20 there. Also, should the acceleration be + or -?

 

[Sassenav22]

acceleration should be positive

 

[Sassenav22]

So what your saying is that the sin20 is suppose to beby the 9.8?

 

[Doc Al]

acceleration should be positive

No. The ramp slopes up, so acceleration is negative (down).

So what your saying is that the sin20 is suppose to beby the 9.8?

Yes, the acceleration of the block is g sin20.

 

[Sassenav22]

Oh, Is the first part of (a) correct because I found the time so that i could find the distance, is that what is suppose to be done?

 

[Doc Al]

Oh, Is the first part of (a) correct because I found the time so that i could find the distance, is that what is suppose to be done?

Yes, that's a perfectly fine way to solve the problem. You just made an error when you did the final arithmetic to solve for the time.

 

[Sassenav22]

so its:

(a) V=u+at
0=5+9.8(sin20)t
t = 1.65s

s= ut+1/2at^2
= 5(1.65)+1/2(-9.8)(sin20)(1.65^2)
=3.69m

(b) v=d/t
5=3.69t
t=3.39/5
t=0.68s

 

[Doc Al]

(a) V=u+at
0=5+9.8(sin20)t
t = 1.65s

The final answer is still wrong. Redo the calculation.

s= ut+1/2at^2
= 5(1.65)+1/2(-9.8)(sin20)(1.65^2)
=3.69m

You'll need to redo this with the correct time.

(b) v=d/t
5=3.69t
t=3.39/5
t=0.68s

No. Read my "tip" in post #2.

 

[Sassenav22]

v=u+at
= 5-9.8(sin20)t
t= 1.64s

I still get the same answer

 

[Doc Al]

v=u+at
= 5-9.8(sin20)t
t= 1.64s

I still get the same answer

I'm curious how. t = 5/(9.8sin20) = ??