Please help with physics? first right answer automatically gets best answer.?

Question

An 90.0 N box of clothes is pulled 23.0 m up a 30.0° ramp by a force of 115 N that points along the ramp. If the coefficient of kinetic friction between the box and ramp is 0.22, calculate the change in the box's kinetic energy.

Just show me how you got it

Answer 1

Let´s make a balance of energy.
Work done on the box =115*23 N m
Increase of potential energy = 90*23*1/2 N m
Work done by the friction forces=90*1/2*sqrt(3)*0.22*23 Nm
Total energy -potential energy -work of friction = kinetic energy
2645-1035-358.53 =1251.47 (check numbers)

Answer 2

Kia ora
First, work is done on the box, because a force is applied over a distance.
Some of the work will be done against friction (this is converted to heat and lost)
Some of the work will be converted into gravitational potential energy
The rest will be kinetic energy.
A)First find the total work done: W=F.d so it's 115*23.0=2,645 J.
B)The height gained is 23 sin30° = 11.5 m. The potential energy gain is mgΔh = 1,035 J
C)The frictional force will be the coefficient of friction multiplied by the normal force = 0.22*mg*cos30° = 17.14N. This is applied over the 23m so the work done against friction is 17.14*23=394.4J.

Now just subtract (B) and (C) from (A) and the balance must be the change in kinetic energy.

Answer 3

So first you should draw a diagram; it really helps.
This question is asking for the change in KE; and we know that W=Change in KE and W=(Net Force)d; So we can set Change in KE EQUAL to (Net Force) d.

-Change in KE= Fd; We are given the distance, and we are missing the Net force applied to the box.

Solving for the Net Force
Along the x-axis (set to be PARALLEL to the ramp's surface) the forces acting on the box are: the Force of KE (Coefficient times the Normal Force), the Weight PARALLEL to the ramp (Weight times the sin of 30), and of course, the Applied force of 115 Newtons. We need to solve for the Normal force to continue, (this will be relatively easy).

Along the y-axis (set to be PERPENDICULAR to the ramp's surface) the forces acting on the box are: the Normal Force and the Weight Perpendicular to the ramp (Weight times the cos of 30). We know that the box is NOT ACCELERATING IN THE Y direction, so given F= 0, the normal force and weight perpendicular have to cancel. W Perpendicular is -77.94 so Normal force has to be +77.94 N.
Thus, we can solve for the net force in the x-direction.

Force of Kinetic Friction: (.22) (Force Normal)= (.22) (77.94N) = -17.147 N

Force of Weight Parallel: (90 N) (sin 30) = -45.0 N
*Note that these forces are negative because they are acting against the applied force of 115 N

And the applied force is given: 115 N.
Thus, summing the forces: 115 N - 17.147 N - 45.0 N= 52.853 N

Finally, solving the question, we can find the change in KE.
W=Fd=Change in KE
(52.853) (distance applied) = Change In KE
(52.853 N) (23 m) = Change in KE

FINAL ANSWER: 1215.619 Joules

Answer 4

If i am reading this right... it sounds like there is no change in mass or velocity.
Since the equation for Kinetic Energy is
KE= (1/2) m v^2
and there is no change in either mass or velocity... the change in KE would be zero. Hope it makes sense

Answer 5

Well, when the box starts moving, using a force diagram I could never draw on here, i calculated the net force of the box moving upwards to be 52.85N.
Method:
.......|\
.mg|.\..mgcos(30)
90N|..\..77.94N
.......|...\
..mgsin(30) 45N

Normal force = mgcos(30) = 77.94
f=mgsin(30)+mu*coef frct
f= 45N + 77.94N*.22
f=62.15N
{F=F+(-f) { stands 4 sum
{F=115N-62.15N
{F=52.85N upwards

Sorry i can't answer the rest of the question yet, we haven't covered how to calculate KE quite yet. I hope what I have told you can help a little though.

Answer 6

Second Newton's law: F + mg + R = ma (vectors).
F - mg sin30° - k mg cos30° = m a
a = [115 - 90(sin30° + 0,22 cos30°)]*9.8/90 = 5.75 m/s^2.
The total work is:
Wtot = [F - mg(sin30°+ 0,22 cos30°)]*d = 1215.6 J
This is the change in kinetic enrgy.