A 3kg package is released at point A on a 51.3degree incline above a long Spring

Question

A 3kg package is released at point A on a 51.3degree incline above a long Spring that is attached as the bottom of the incline. The Spring constant is 140Nm and the top end of the Spring is B.(distance=4m). The friction between the package and Incline is 0.2. The mass of the Spring is negligible.

a) What is the normal force acting on the package by the incline?

b) What is the friction force acting on the package by the incline?

c) What is the work done by the friction on the package just before it reaches the Spring?

d) What is the change of the graviational potential energy of the package?

e) What is the Kinetic Energy of the package just before it reaches the Spring at point B?

f) What is the speed of the package just it reaches the Spring at point B?

f) What is the maximum tension of the spring?

For the final answer please add a daigram if you can. Thank you.

Explanation / Answer

a) mgcos(51.3) =18.400891376 N perpendicular to incline

b) 0.2 * mgcos(51.3) => 3.68017827519 N

c) work = Force * distance = 0.2 * mgcos(51.3) * 4 => 14.7207131008 J

d) mgsin(51.3) * 4 (distance of 4 meter on incline means 4*sin(51.3) in height ) = 91.8722675519 N

e) KE =Work done by all forces = mgsin(51.3) * 4 (BY gravity) - 0.2 * mgcos(51.3) * 4 (BY friction)

=25.717184817 J

f) 25.717184817 = 1/2*m*v^2

v= 12.4218802483 m/s

g) since all of kinetic energy is converted to spring energyso

25.717184817 = 1/2* 140 * x^2

0.15127755774 = x^2

x= 0.38894415761 m

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