Applying Newton's Laws Projected Up an Incline

[djester555]

Projected Up an Incline A block is projected up a frictionless inclined plane with initial speed v1 = 3.43 m/s. The angle of incline is θ = 32.5°.
(a) How far up the plane does it go?

(b) How long does it take to get there?

(c) What is its speed when it gets back to the bottom?



What i got soo far
v * T = d

a) 3.43 * .674 = 2.31 m


b)vi = 3.43
vf = 0
a = - 9.80 sin32.5
t = ?

t = (vf - vi) / a
t = 3.43 / (9.80 sin32.5)
t = 0.674




c) don't even know where to begin on this one

 

[wavingerwin]

You can solve the problem using the equations of motion

vf² = vi² +2ad ---------(1)
vf = vi + at --------(2)
where vf and vi are the final and initial velocity respectively and a is the acceleration, t is the time

Using equation (1), you can solve for d
and using (2) you can solve for t

For the third question, you have to consider the distance the block has traveled
and the block would have this length of 'track' to go back to the bottom.

So, again you can use (1) to solve for vf
(which must equal to 3.43m/s, to the opposite direction)

i don't get what is 'T' used for question a
your working for b is correct but i get 0.651s as my answer, using the same working

 

[tomcue]

To calculate the speed of the block when it gets back to the bottom, we can use the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. We already know the initial velocity (3.43 m/s) and the time it takes to reach the top (0.674 seconds), so we just need to calculate the acceleration.

To do this, we can use Newton's second law, which states that force (F) equals mass (m) multiplied by acceleration (a). In this case, the only force acting on the block is gravity, which can be calculated using the formula F = mg, where m is the mass of the block and g is the acceleration due to gravity (9.80 m/s^2).

So, we can set up the equation F = ma and substitute in the known values, giving us mg = ma. We can then cancel out the mass on both sides, leaving us with a = g.

Now, we can plug in the value for acceleration (9.80 m/s^2) and the time it takes to reach the top (0.674 seconds) into the equation vf = vi + at to calculate the final velocity.

vf = 3.43 + (9.80)(0.674)
vf = 9.34 m/s

Therefore, the speed of the block when it gets back to the bottom is 9.34 m/s.