Kinetic Friction with an upward shove

[lauriecherie]

Homework Statement

In the figure below, the coefficient of kinetic friction between the block and inclined plane is 0.17 and angle θ = 55°. What is the block's acceleration (magnitude and direction) assuming the following conditions? (b) It has been given an upward shove and is still sliding up the slope.
I also figured out that the accelertaion (magnitude and direction) when it is sliding down. That answer is 7.079 m/s^2

Homework Equations

The Attempt at a Solution

I've tried 8.036 & 7.079 m/s^2. Both were incorrect :(

 

[alphysicist]

Hi lauriecherie,

Homework Statement

In the figure below, the coefficient of kinetic friction between the block and inclined plane is 0.17 and angle θ = 55°. What is the block's acceleration (magnitude and direction) assuming the following conditions? (b) It has been given an upward shove and is still sliding up the slope.
I also figured out that the accelertaion (magnitude and direction) when it is sliding down. That answer is 7.079 m/s^2

Homework Equations

The Attempt at a Solution

I've tried 8.036 & 7.079 m/s^2. Both were incorrect :(

What did you do to get those answers?

 

[nlightner]

I would first clarify the given information. The coefficient of kinetic friction between the block and inclined plane is 0.17 and the angle of the inclined plane is 55°. The block has been given an upward shove and is still sliding up the slope. The question asks for the block's acceleration in terms of magnitude and direction.

To solve this problem, we can use the equation for Newton's second law of motion, F=ma, where F is the net force acting on the block, m is the mass of the block, and a is the acceleration of the block.

First, we need to determine the net force acting on the block. Since the block is sliding up the slope, the net force must be directed up the slope. The only forces acting on the block are the force of gravity (mg) and the force of kinetic friction (μkN), where N is the normal force exerted by the inclined plane on the block.

We can find the normal force by breaking it into its components. The normal force is perpendicular to the inclined plane, so its vertical component is equal in magnitude and opposite in direction to the force of gravity (mgcosθ). The horizontal component is equal to the force of gravity acting down the slope (mgsinθ).

Next, we can plug these values into the equation for the net force, F=ma. Since the block is still sliding up the slope, the acceleration must be directed up the slope as well. Thus, we can set up the following equation:

ma = μkN - mgsinθ

Substituting in the values for N and sinθ, we get:

ma = μk(mgcosθ) - mg(sinθ)

Solving for a, we get:

a = (μkcosθ - sinθ)g

Plugging in the given values of μk = 0.17 and θ = 55°, we get:

a = (0.17cos55° - sin55°)(9.8 m/s^2) = 1.95 m/s^2

Therefore, the block's acceleration in terms of magnitude and direction when it is sliding up the slope is 1.95 m/s^2 directed up the slope. This is significantly lower than the acceleration when the block is sliding down the slope, which is 7.079 m/s^2 directed down the slope. This is due to the opposing