A 2 kg package is released from rest on a 53.1 degree incline 4 meters from a sp

Question

A 2 kg package is released from rest on a 53.1 degree incline 4 meters from a spring of force constant 120 N/m. The coefficient of kinetic friction between the block and surface is 0.20. Write out the energy equation between the point when first released and the moment the block hits the spring. m = 2 00 Write out the energy equation between the points when first released and when the spring is at maximum compression. If m is small enough, the block will rebound back up the incline 2 meters past the unstretched spring. Write out the energy equation 4 00 m' between the points when first released and the moment the block rebounds to its maximum height.

Explanation / Answer

a)

l = 4 m

at the first rrelease the block has potential energy U1 = m*g*h

h = vertical height of the block from the horizantal surface = l*sin53.1

at the moment htting the spring KE1 = 0.5*m*v^2

work done by frictional force = Wf = uk*m*g*l*cos53.1

work energy relation

U1 - K1 = Wf

m*g*l*sin53.1 - 0.5*m*v^2 = uk*m*g*l*cos53.1

9.8*4*sin53.1 - 0.5*v^2 = 0.2*9.8*4*cos53.1

v = 7.3 m/s

(b)

let the spring compresses a distance x

iniital potential energy = U1 = m*g*(i+x)*sin53.1

Wf = uk*m*g*(l+x)*cos53.1

after the spring is compresses speed of the block = 0

KE = K1 = 0

potentaial energy stored i spring = U2 = 0.5*k*x^2

from work energy relation

U1 - U2 = Wf

m*g*(l+x)*sin53.1 - 0.5*k*x^2 = uk*m*g*(l+x)

(2*9.8*(4+x)*sin53.1) - (0.5*120*x^2) = (0.2*2*9.8*x)

x = 1.12 m

a)

l = 4 m

at the first rrelease the block has potential energy U1 = m*g*h

h = vertical height of the block from the horizantal surface = l*sin53.1

at the moment htting the spring KE1 = 0.5*m*v^2

work done by frictional force = Wf = uk*m*g*l*cos53.1

work energy relation

U1 - K1 = Wf

m*g*l*sin53.1 - 0.5*m*v^2 = us*m*g*l*cos53.1

(b)

let the spring compresses a distance x

iniital potential energy = U1 = m*g*(i+x)*sin53.1

Wf = us*m*g*(l+x)*cos53.1

after the spring is compresses speed of the block = 0

KE = K1 = 0

potentaial energy stored i spring = U2 = 0.5*k*x^2

from work energy relation

U1 - U2 = Wf

m*g*(l+x)*sin53.1 - 0.5*k*x^2 = us*m*g*(l+x)

(c)

initial potential energy U1 = m*g*2*sin53.1

work done by friction = us*m*g*6*cos53.1

final potnetial energy = 0

U1 = Wf

m*g*2*sin53.1 = us*m*g*6*cos53.1

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