could you help me its due at midnight and ill fail if somone doesn't come through?

Question

An 75.0 N box of clothes is pulled 16.0 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.

Show me how to plug in the numbers and what you did to find fnet

Answer 1

I did this in a different approach than the person above.
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 -64.95 N so Normal force has to be 64.95 N.
Thus, we can solve for the net force in the x-direction.

Force of Kinetic Friction: (.22) (Force Normal)= (.22) (64.95 N) = -14.29 N

Force of Weight Parallel: (75 N) (sin 30) = -37.5 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 - 14.29 N - 37.5 N= 63.21 N

Finally, solving the question, we can find the change in KE.
W=Fd=Change in KE
(63.21 N) (distance applied) = Change In KE
(63.21 N) (16 m) = Change in KE

FINAL ANSWER: 1011.36 Joules

Answer 2

Given:
W = 75.0 N, weight of box of clothes
dx = 16.0 m, distance traveled upward along the ramp by box
θ = 30 deg, angle of inclination of the ramp
Fa = 115 N, applied force parallel to the ramp
μk = 0.22, coefficient of kinetic friction between box and ramp

Find:
ΔK.E. = change in kinetic energy of the box

Solution:

ΔK.E. = K.E.2 - K.E.1
ΔK.E. = (1/2)(m)vf^2 - (1/2)(m)vi^2
ΔK.E. = (1/2)(W/g)(vf^2) - (1/2)(W/g)(0)^2
ΔK.E. = (1/2)(75.0/9.8)(vf^2)
ΔK.E. = 3.83(vf^2) equation 1

Fnet = ma
Fa - Ff - Wx = (W/g)(a), Wx is the component of W along the x-axis, which is Wx = Wsinθ
115 N - μk(Fn) - Wsinθ = (75/9.8)(a)
115 N - 0.22(Fn) - (75/9.8)sin 30.0 deg = 75/9.8(a)
Fn = Wy
Fn = Wcosθ
Fn = 75.0N cos 30.0 deg
Fn = 64.95 N
115 N - 0.22(Fn) - (75/9.8)sin 30.0 deg = 75/9.8(a)
115 N - 0.22(64.95) - (75/9.8)sin30.0 deg = 7.65(a)
96.9 = 7.65a
a = 96.9/7.65
a = 12.7 m/s^2

vf^2 - vi^2 = 2adx
vf^2 - 0 = 2adx
vf = √(2adx)
vf = √(2)(12.7)(16.0)
vf = 20.2 m/s

Therefore, ΔK.E. = 3.83(vf^2) equation 1
ΔK.E. = 3.83(20.2^2)
ΔK.E. = 1562.8 Joules ANS

teddy boy

Answer 3

Work done was wrong N m(should be115*16=1840)
Increase of potential energy =16*1/2*75=600N m
Work done by friction = 75*1/2*sqrt3 *16*0.22=171.5 Nm
Increase of kinetic energy = 1840-600-171.5=1068.5 Joule

Answer 4

net force (up, along) = m*a = F(up,along) - mg sin 30 (down, along) - friction force (down. along)
ma = F - mg sin 30 - u [mg cos 30]
ma = 115 - 75*0.5 - 0.22 [75*0.866] = 63.211 Newton
-----------------------------
so we have net acceleration up along
V (final)^2 = u^2 + 2 * a * L
m[v^2 - u^2] = 2 m a L = 2L * 63.211 ---(2)
==========
Increase in KE = 0.5m[v^2 - u^2] = L * 63.211 = 16*63.211
= 1011.38 Joule