Two Physics Questions (again)?

Question

--An 80.0 N box of clothes is pulled 22.0 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.

-- A 103.0 N grocery cart is pushed 12.0 m along an aisle by a shopper who exerts a constant horizontal force of 40.0 N. If all frictional forces are neglected and the cart starts from rest, what is the grocery cart's final speed?

i've done some work for each but the online worksheet keeps on telling me my answers are wrong. could somebody please step by step write out how to do these problems (w/ numerical values if possible)

Answer 1

Hey Big Sur,

1. gp4rts started out good but, if I understand what he said, I disagree with him toward the end.

The 115 N does work that needs to be accounted for.
Work in = work by friction + gain in potential energy + gain in KE

Work in = F*d = 115 N*22 m
work by friction = u*N*d = 0.22*80.0 N*cos(30)*22.0 m
gain in potential energy = w*d*sin(30) = 80.0 N*22.0 m*sin(30)
gain in KE = ?

gain in KE = Work in - work by friction - gain in potential energy
I get
gain in KE = 2530 - 335.3 - 880 = 1315 joules

2. You can do this with the equations of motion after finding the mass and then acceleration. Or use the conservation of energy. I'll do equations for both
motion:
m = w/g, a = F/m, V^2 = Vo^2 + 2*a*d
energy
work in = KE
F*d = (1/2)*m*V^2

Say hi to an otter for me

Answer 2

Are you sure you are taking into account all of the energy? The change in potential energy is W*h, where W is the weight and h is the altitude. On a ramp, that will be h=L*sinø, where L is the distance up the ramp that the box is moved. The energy taken by friction is F*L, where F is the frictional force. But F is = µ*Fn, where Fn is the component of F normal to the ramp; this is W*cosø. µ is the coeff of friction. The change in kinetic energy will be the sum of the change in potential energy and the energy taken by friction. Note that the force pushing the box does not enter into the calculation. That force only determines how fast the energy will change, but the amount of energy change is the same no matter how long it takes for the box to get to the end of the ramp.

The cart's position under constant acceleration is given by s =.5*a*t^2. It's final speed is V=a*t. From Newtons law you know that a=F/m. (You are given the weight in newtons. Divide by g to get the mass.) You are given s. Solve for t from the first equation, an put that value in the second equation to get V. Or you can make the substitution algebraically to get v = √[2*a*s]

EDIT: sojail says "work by friction = u*N*d". That is not correct. The frictional force is not u*N, but the normal (to the ramp) component of weight times u.

Note: The ACTUAL kinetic energy will depend on the force, but the CHANGE does not.