# Incline Question

#### niel23

2+ Year Member
A block is on an inclined plane with a string attached from the block to a pulley at the top of the incline.
The coefficient of kinetic friction between the mass and the plane is 0.1. The mass is 100 kg. The plane is inclined at 30 degrees.

If the mass is moving up the inclined plane at a velocity of 2 m/s, what tension should be applied to the rope in order to make the mass reverse directions in exactly 1 s?

My method of solving this:

I first set up my equation as:
(force tension) - (force gravity x) - (force friction) = ma

Since acceleration will abruptly change in the downward direction, solving for a:
v final = v + at
0 = 1 + a (1 second)
a = -1 m/s^2 ---> since its going down the slope, sign will change to +

Plugging back into first question and solving for force of tension

force tension = [100 kg (10 m/s^2) sin 30] + [0.1 (100)(10) cos 30] + 100 (1 m/s^2)
Ft = 685 N

However, this is incorrect. The answer is 387 N.

I'm stumped how to solve this problem. I'd appreciate any help on this, thanks!

#### theonlytycrane

7+ Year Member
Do you mean the block is sliding down the plane at 2 m/s?

Do you want the block to be held in place or then be pulled up the plane at 2 m/s?

#### betterfuture

2+ Year Member
Since velocity changed, there is an acceleration. If the mass experiences an acceleration, there is a force.

Fnet=ma=T-fk+mgsin30degrees

Since the velocity initially was 2m/s and it reversed direction, the instant it did, its final velocity became 0.

Using vf=vo+at, solve for a. Isolate T by itself after figuring out the net Force in the x direction (Fnet=100kg*acceleration) and solve.

#### niel23

2+ Year Member
@theonlytycrane, the block is sliding up the plane with an initial velocity of 2 m/s. However, after 1 sec, the block reverses direction and slides back down.

@betterfuture, my thought pattern followed that, but when solving it out, the answer I calculated was greater, 685N.

#### betterfuture

2+ Year Member
You set up the equation incorrectly.

(force tension) - (force gravity x) - (force friction) = ma

The tension is opposite the force of gravity and friction so it would be
Fnet=ma=T-fk+mgsin30degrees

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