A red brick with mass m=4.5 kg is launched over terrain defined by heights h1=2.

Q: A red brick with mass m=4.5 kg is launched over terrain defined by heights h1=2.

use the law of conswrvation of energy as PE = EPE mgh = 0.5 kx^2 4.5 * 9.8 *(2.5-2) = 0.5*22e 3 *x^2 x = 4.4 cm -------------------------------- Net Force=Frictional force ma = uk mg ma = uk mg a= 0.4 * 4.5 * 9.8 /4.5 a = 3.92 m/s^2 use 0.5 mv^2 = 0.5 kx^2 v^2 = 22 e 3 * 0.044* 0.044/(4.5) v = 3.07 m/s so Distance travelled S = V^2/2a S = 3.07* 3.07/(2* 3.92) S = 1.202 m

ID: 1345110 • Letter: A

Question

A red brick with mass m=4.5 kg is launched over terrain defined by heights h1=2.5 m and h2=2 m by an ideal spring as shown in Fig. 2. The spring constant is k=22. 103 N/m. The first part of the path is frictionless but on the horizontal part after the hill there is friction that can be described by a kinetic coefficient of friction µk=0.4.

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a) The brick is launched with this minimal spring compression. Calculate how far along the horizontal friction track the brick moves before coming to rest.

Explanation / Answer

use the law of conswrvation of energy as PE = EPE

mgh = 0.5 kx^2

4.5 * 9.8 *(2.5-2) = 0.5*22e 3 *x^2

x = 4.4 cm

--------------------------------
Net Force=Frictional force

ma = uk mg

a= 0.4 * 4.5 * 9.8 /4.5

a = 3.92 m/s^2

use 0.5 mv^2 = 0.5 kx^2

v^2 = 22 e 3 * 0.044* 0.044/(4.5)

v = 3.07 m/s

so Distance travelled S = V^2/2a

S = 3.07* 3.07/(2* 3.92)

S = 1.202 m

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A red brick with mass m=4.5 kg is launched over terrain defined by heights h1=2.

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A red brick with mass m=4.5 kg is launched over terrain defined by heights h1=2.

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