Spring with a spring constant of 400 N/m is used to propel a 0.500-kg mass up an

Question

Spring with a spring constant of 400 N/m is used to propel a 0.500-kg mass up an inclined plane. The spring is compressed 0.3 m from its equilibrium position and the mass from rest across a horizontal surface and onto the plane. The plane has a length of 400 m ans is inclined at angle theta = 30.0 degree. The coefficient of kinetic friction between the mass and the horizontal surface is 0.4, but the plane is frictionless. When the spring is compressed, the mass is 1.50 m from the bottom of the plane as shown in the figure. a) What is the work done by friction on the mass as it moves across the horizontal surface? b) What is the kinetic energy of the mass as it reaches the bottom of the plane? c) What is the speed of the mass as it reaches the top of the plane?

Explanation / Answer

here,

spring constant , k = 400 N/m

mass , m = 0.5 kg

compression in the spring , x = 0.3 m

length of plane , l = 4 m

theta = 30 degree

uk = 0.4

l1 = 1.5 m

a)

the work done by friction , Wf = - uk * m * g * l1

Wf = - 0.4 * 0.5 * 9.8 * 1.5 J

Wf = - 2.94 J

b)

the kinetic energy of the mass as it reaches the bottom of the plane , KE = potential energy stored in the spring + work done by friction

KE = 0.5 * k * x^2 + Wf

KE = 0.5 * 400 * 0.3^2 - 2.94

KE = 15.06 J

c)

let the speed of the mass at the top of the plane be v

using conservation of energy

KE = KEf + PEf

15.06 = 0.5 * m * v^2 + m * g * l * sin(theta)

15.06 = 0.5 * 0.5 * v^2 + 0.5 * 9.8 * 4 * sin(30)

solving for v

v = 4.59 m/s

the speed at the top of the incline is 4.59 m/s

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