Question

A freight company uses a compressed spring to shoot 1.90kg packages up a

1.00-m -high frictionless ramp into a truck, as the figure(Figure 1) shows. The spring constant is 330N/m and the spring is compressed 35.0cm .

What is the speed of the package when it reaches the truck?

Express your answer with the appropriate units.

v =

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Part B

A careless worker spills his soda on the ramp. This creates a 50.0-cm-long sticky spot with a coefficient of kinetic friction 0.300. Will the next package make it into the truck?

A careless worker spills his soda on the ramp. This creates a 50.0--long sticky spot with a coefficient of kinetic friction 0.300. Will the next package make it into the truck?

	Yes
	No

Answer 1

k be the spring constant,

x be the compression of the spring,

v0 be the initial speed of the package,

v1 be the final speed of the package,

m be the mass of the package,

R be the normal reaction of the slope on the package,

h be the vertical height of the ramp,

L be the length of the sticky patch,

F be the friction force down the plane,

a be the angle of inclination of the plane,

u be the coefficient of kinetic friction,

g be the acceleration due to gravity.

1.

Equating the energy in the spring to the initial KE of the package:

kx^2 / 2 = mv0^2 / 2 .

kx^2 = mv0^2 ..(1)

Equating initial KE of the package to KE plus PE on arrival at the truck:

mv0^2 / 2 = mgh + mv1^2 / 2

v1^2 = v0^2 - 2gh ...(2)

Eliminating v0^2 from (1) and (2):

kx^2 = m(v1^2 + 2gh) ...(3)

v1^2 = kx^2 / m - 2gh

v = sqrt[ kx^2 / m - 2gh ]

= sqrt[ 328 * 0.35^2 / 2.00 - 2 * 9.81 * 1.00 ]

= 0.686 m/s.

2.

If R is the normal reaction of the ramp on the package:

mg cos(a) = R

The friction force resulting from this is:

F = uR

= umg cos(a) parallel to the plane.

The energy lost as a result is:

umgL cos(a)

Incorporating this into (3):

kx^2 - umgL cos(a) = m(v1^2 + 2gh)

kx^2 / m - ugL cos(a) = v1^2 + 2gh

v1^2 = kx^2 / m - g[ 2h - uL cos(a) ]