Attachment 25974 answers are in the image i just dont get the third part could anyone just solve it ?
Printable View
Attachment 25974 answers are in the image i just dont get the third part could anyone just solve it ?
Oh! You want some help. Please show some work next time, or at least your best guess.
For part iii. You know the speed of the block up the incline from part ii. Now this is just a motion in 1-D problem. You have the initial speed of the block (from part ii) and you are looking for the time such that the block comes to rest on the incline. The only thing you need is an acceleration, and you can find by finding the component of the weight down the incline. (Divide by m and you get the acceleration on the block.)
Give it a try from here and let us know how you are doing on it.
-Dan
i use V=U+at as we dont know the distance covered so this is the only equation without distance. since v=0 and a=g the final velocity of part (ii) is the initial velocity of part (iii) . the problem is i get the negative answer as -U=at now can u help
a is not equal to g. Notice that the mass is on an incline. So you need to find the component of the weight (which is straight down) along the incline. I don't do pictures well so I'll give you a description of how to find it. You have your sketch of the problem. Where the mass is I want you to draw an arrow representing the Normal force. (The direction will be perpendicular to the incline.) The weight is pointed straight down, so sketch that vector. (Make it a little long for convenience.) Now extend the line of the normal force downward. (That is perpendicular to the incline, but in the opposite direction of the normal force.) Using this line draw a new line at right angles from the line you just drew until it meets the bottom of your weight line.
Notice that this last line is pointing down the incline. This is the weight component in the direction of the incline. Now. The angle of the incline is the same as the angle between the weight vector and the line pointing in the opposite direction of the Normal line.
I have included the diagram which should (hopefully!) make the text make some more sense.
So assuming you've got the acceleration down the slope...Call up the slope positive. So your initial speed is positive (up the slope) and the acceleration is negative (down the slope.) You've got the initial speed, the final speed (0 m/s) and the acceleration. You are looking for the time where the object stops. What equation has all of these? Looks to me like you need.
-Dan
surprisingly enough i got the answer the acceleration is a= -g after the M hits the floor well i got the answer thanks anyways