How is Work Done on an Object Thrown Up an Inclined Plane Calculated?

[Felafel]

Homework Statement

An object, (mass=15 kg) is thrown up a 30° inclined plane, with initial velocity=4.6 m/s
The coefficient of friction is μ=0.34. Find the work done on the object by the normal force, the resultant force, the weight and the friction, from the beginning until it stops (so not when it slides back). Which is the distance run by the object?

The Attempt at a Solution

Could anyone please check it? Because I think the results are correct, but I don't know anything about the force which throws the object and I've solved all without taking it into account.

The work of the normal force is 0 because the angle between it and the direction of the motion is 90°

Then I use the formula $\Delta K=W_{ris}$ with $K_f = 0$ because $v_f=0$
so $W_{ris}=\frac{1}{2}\cdot m \cdot v_i^2=159J$

The work of the resultant force is equal to the work of the friction + the work of the weight
And

$F_k (friction)=m \cdot g \cdot cos30° \cdot μ$ = 43 N
$F_w=m \cdot g \cdot sin30°=74 N$
$F_{ris}=F_w+F_k=117N$
So
$W_{ris}=F_{ris} \cdot r \cdot cos0° \Rightarrow r=1.4 m$

$W_w= r \cdot \ cos180° \cdot F_w = - 104 J$
$W_k= r \cdot \ cos180° \cdot F_k= -60 J$

thank you :)

 

[BruceW]

yep. Very good, all looks correct to me.

Edit: and yes, you are not supposed to take into account the force with which the object is thrown. The idea is that the initial velocity of 4.6 m/s is its velocity after any initial 'throwing force' has been applied to it. (so the initial velocity contains all the information to do with how hard it was thrown).

 

[Felafel]

:D great! thanks again

 

[BruceW]

no worries :) glad I could help (or at least confirm, haha)